---
title: "Power Series"
book: "University Mathematics — Year 2"
subject: math
language: en
chapter: 11
exercises: 12
source: https://one-course.com/books/math/4/en/chapter/11-power-series
---

# Chapter 11 — Power Series

Power series are the best-behaved series of functions in mathematics: inside their disk of convergence they converge [normally](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) on [compacts](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact), may be differentiated and integrated term by term without a second thought, and their sums — *[analytic functions](#def-b2-powerseries-analytic)* — are determined by their coefficients. This chapter proves that whole package and reclaims, honestly, every Taylor series of the Year 1 volume; [generating functions](#ex-b2-powerseries-fibonacci) close it with algebraic dividends.

## 11.1 Radius of convergence

**Lemma 11.1 (Abel).**

If the sequence $(a_n z_0^n)$ is bounded for some $z_0 \neq 0$, then $\sum a_n z^n$ converges [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) for every $\abs z < \abs
{z_0}$, and [normally](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) on every disk $\abs z \leq r < \abs{z_0}$.

**Proof.** With $\abs{a_n z_0^n} \leq M$ and $\abs z \leq r$:

$$
\abs{a_n z^n} = \abs{a_n z_0^n}\,\Bigl|\frac{z}{z_0}\Bigr|^n
\leq M\Bigl(\frac{r}{\abs{z_0}}\Bigr)^{\!n},
$$

a convergent geometric bound, uniform on the disk. ∎

**Definition 11.2 (Radius of convergence).**

The *radius of convergence* of $\sum a_n z^n$ is

$$
R = \sup\{r \geq 0 : (a_n r^n) \text{ bounded}\} \in
\intcc{0}{+\infty} .
$$

By [Lemma 11.1](#lem-b2-powerseries-abel): absolute convergence for $\abs
z < R$ (normal on [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) subdisks), divergence — indeed unbounded terms — for $\abs z > R$. On the boundary circle, anything can happen ([Exercise 11.2](#exo-b2-powerseries-2)). In practice $R$ is computed by d’Alembert’s ratio test on $\abs{a_n}\abs z^n$ or by comparison.

**Example 11.3 (A radius with no ratio test).**

What is the radius of $\sum \sin(n)\,z^n$? The ratio $\abs{\sin(n+1)/\sin n}$ has no limit, but the definition works directly. *$R \geq 1$:* $\abs{\sin n} \leq 1$, so $(\sin
n\cdot r^n)$ is bounded for every $r < 1$ — indeed for $r =
1$. *$R \leq 1$:* it suffices that $\sin n \not\to 0$. Suppose $\sin n \to 0$; the addition formula

$$
\sin(n+1) = \sin n\cos 1 + \cos n\sin 1
$$

would force $\cos n \to 0$ (solve for $\cos n$: $\sin 1 \neq
0$), contradicting $\sin^2 n + \cos^2 n = 1$. So the terms $\sin(n)\,1^n$ do not tend to $0$: the series diverges at $z =
1$, and $R \leq 1$. Conclusion: $R = 1$. Closing insight: the radius is a statement about $\abs{a_n}r^n$ being bounded — no limit of ratios is ever required, and boundedness arguments settle cases the ratio test cannot touch (compare the oscillating coefficients of [Exercise 11.1](#exo-b2-powerseries-1)).

**Proposition 11.4 (Operations).**

Let $\sum a_nz^n$, $\sum b_nz^n$ have radii $R_a, R_b$. Then, for $\abs z < \min(R_a, R_b)$:

$$
\sum (a_n + b_n)z^n = \sum a_nz^n + \sum b_nz^n,
\qquad
\Bigl(\sum a_nz^n\Bigr)\Bigl(\sum b_nz^n\Bigr) = \sum c_n z^n,
\quad c_n = \sum_{k=0}^{n} a_kb_{n-k},
$$

both series having radius $\geq \min(R_a, R_b)$. (The product is the [Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini), legitimate by absolute convergence and [Theorem 7.14](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini).)

**Proof.** The sum formula is the linearity of convergent series, and $(a_n + b_n)r^n$ is bounded whenever both $a_nr^n$ and $b_nr^n$ are: radius $\geq \min(R_a, R_b)$. For the product, fix $\abs z
< \min(R_a, R_b)$: both series converge *[absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def)* there ([Lemma 11.1](#lem-b2-powerseries-abel)), so the doubly indexed family $(a_kz^k\,b_lz^l)_{k,l}$ is [summable](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-summable), and [Theorem 7.14](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini) allows any grouping. Grouping by $k
+ l = n$:

$$
\Bigl(\sum_k a_kz^k\Bigr)\Bigl(\sum_l b_lz^l\Bigr)
= \sum_{n\geq0}\Bigl(\sum_{k+l=n}a_kb_l\Bigr)z^n
= \sum_{n\geq0}c_nz^n ,
$$

[absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) convergent for every such $z$: the product series has radius $\geq \min(R_a, R_b)$ as well. ∎

**Example 11.5 (A Cauchy square, cross-checked).**

Square the geometric series: for $\abs x < 1$, the coefficient of $x^n$ in $\bigl(\sum x^k\bigr)^2$ is $c_n = \sum_{k+l=n}
1\cdot1 = n + 1$, so

$$
\frac{1}{(1-x)^2} = \sum_{n\geq0}(n+1)\,x^n .
$$

Cross-check by term-by-term differentiation ([Theorem 11.7](#thm-b2-powerseries-calculus) below): differentiating $\frac{1}{1-x} = \sum x^n$ gives $\frac{1}{(1-x)^2} = \sum
nx^{n-1} = \sum(n+1)x^n$ — the same series by two unrelated mechanisms. Closing insight: when a coefficient identity looks mysterious, one of these two engines (convolution or differentiation) usually produces it in one line; the weekend problem’s question on $\sum\binom{2k}k\binom{2n-2k}{n-k} = 4^n$ runs the convolution engine at full power.

**Example 11.6 (Multiplying by 11−x\frac{1}{1-x}1−x1​ sums the coefficients).**

A [Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini) against the geometric series has a memorable meaning: for any $\sum a_nx^n$ of radius $R > 0$ and $\abs x <
\min(R, 1)$,

$$
\frac{1}{1-x}\sum_{n\geq0}a_nx^n
= \sum_{n\geq0}\Bigl(\sum_{k=0}^{n}a_k\Bigr)x^n :
$$

multiplication by $\frac{1}{1-x}$ replaces coefficients by their partial sums (convolve with the all-ones sequence). Instance: $\dfrac{\eu^x}{1-x} = \sum_n s_n x^n$ with $s_n =
\sum_{k\leq n}\frac{1}{k!}$, the partial sums of $\eu$ — compare [Exercise 11.11](#exo-b2-powerseries-11), where the same product with $\eu^{-x}$ encodes the derangement counts. Closing insight: operations on power series are operations on coefficient sequences in disguise (multiply by $\frac1{1-x}$: sum; multiply by $x$: shift; differentiate: multiply by $n$ and shift) — a dictionary the generating-functions chapter will read fluently.

## 11.2 Regularity of the sum

**Theorem 11.7 (Term-by-term calculus).**

Let $f(x) = \sum_{n\geq0} a_n x^n$ have radius $R > 0$ (real variable $x \in \intoo{-R}{R}$).

1. The derived series $\sum n\,a_n x^{n-1}$ has the *same* radius $R$, and $f$ is $C^1$ with $f'(x) = \sum_{n \geq 1}  n a_n x^{n-1}$. Iterating, $f$ is $C^\infty$ and $$a_n = \frac{f^{(n)}(0)}{n!} :$$ the coefficients of a power series are unique (two series with the same sum near $0$ have equal coefficients).
2. Term-by-term primitive: $\sum \frac{a_n}{n+1}x^{n+1}$ has radius $R$ and derivative $f$ .

**Proof.** *Same radius:* if $(a_nr^n)$ is bounded and $r' < r$, then $n\abs{a_n} r'^{\,n-1} = \frac{n}{r'}\abs{a_nr^n}
\bigl(\frac{r'}{r}\bigr)^n$ is bounded (indeed $\to 0$: geometric beats $n$), so $R' \geq R$; conversely $\abs{a_n x^n} \leq \abs x
\cdot n\abs{a_n}\abs x^{n-1}$ gives $R \geq R'$.

*Differentiation:* on $\intcc{-r}{r}$, $r < R$, the derived series converges [normally](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) ($n\abs{a_n}r^{n-1}$ [summable](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-summable) by the radius computation); the original converges at $x = 0$: the differentiation theorem for series ([Theorem 10.11](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#thm-b2-funcseq-seriestransfer)) applies on every such segment, hence on $\intoo{-R}{R}$. Iterating $k$ times and evaluating at $0$: explicitly, the $k$-th derived series is

$$
f^{(k)}(x) = \sum_{n\geq k} n(n-1)\cdots(n-k+1)\,a_n\,x^{n-k},
$$

and at $x = 0$ every term with $n > k$ vanishes, leaving only the constant term $k(k-1)\cdots1\cdot a_k$: $f^{(k)}(0) =
k!\,a_k$. Uniqueness of coefficients follows: two power series with the same sum near $0$ have the same derivatives at $0$, hence the same $a_k$. Primitives: same radius by the same computation, and differentiate term by term back. ∎

**Example 11.8 (Evaluating a series at a point).**

What is $\sum_{n\geq1}\dfrac{n^2}{2^n}$? It is the sum $\sum n^2x^n$ of [Exercise 11.3](#exo-b2-powerseries-3) evaluated *inside* the disk, at $x = \frac12 < 1 = R$, where every manipulation used to derive the closed form was legitimate:

$$
\sum_{n\geq1} n^2x^n = \frac{x(1+x)}{(1-x)^3}
\quad\Longrightarrow\quad
\sum_{n\geq1}\frac{n^2}{2^n}
= \frac{\frac12\cdot\frac32}{(\frac12)^3}
= \frac{3/4}{1/8} = 6 .
$$

Same engine, other dials: $x = \frac13$ gives $\sum\frac{n^2}{3^n} = \frac{\frac13\cdot\frac43}{(2/3)^3} =
\frac32$. Closing insight: a power-series identity is a machine, not a single formula — one derivation prices every numerical series $\sum n^2q^n$ at once, for all $\abs q < 1$; this is how the generating-functions chapter will compute expectations and variances wholesale.

**Example 11.9 (The classics, this time honestly).**

$\displaystyle\frac{1}{1 - x} = \sum x^n$ ($R = 1$); integrating term by term ([Theorem 11.7](#thm-b2-powerseries-calculus) (2)):

$$
-\ln(1 - x) = \sum_{n\geq1} \frac{x^n}{n},
\qquad
\arctan x = \sum_{n \geq 0} \frac{(-1)^n x^{2n+1}}{2n+1}
\quad (\abs x < 1),
$$

the second in two steps: substitute $-x^2$ in the geometric series to get $\frac{1}{1+x^2} = \sum(-1)^nx^{2n}$ (radius $1$, since $\abs{x^2} < 1 \iff \abs x < 1$), then take the term-by-term primitive vanishing at $0$; both sides are primitives of the same function with the same value at $0$, hence equal on $\intoo{-1}{1}$. And $\exp$: the series $E(x) = \sum \frac{x^n}{n!}$ ($R = \infty$) satisfies $E' = E$, $E(0) = 1$ by termwise differentiation, so $E = \exp$ by the Year 1 uniqueness. Every “standard expansion” of the Year 1 volume is now a theorem about its full power series.

**Example 11.10 (A logarithm computed from inside the disk).**

Evaluating $-\ln(1-x) = \sum\frac{x^n}{n}$ at the interior point $x = \frac12$:

$$
\sum_{n\geq1}\frac{1}{n\,2^n} = \ln 2 ,
$$

a fast-converging representation of $\ln 2$ (ten terms already give $0.69306\ldots$ against $\ln 2 = 0.69314\ldots$), far better than the alternating series $1 - \frac12 + \frac13 -
\dots$ available only at the boundary. Closing insight: whenever a constant is reachable both at the rim and strictly inside the disk, the interior wins numerically — geometric decay against harmonic decay.

**Example 11.11 (Differentiation preserves the radius, not the boundary).**

The series $\sum_{n\geq1}\frac{x^n}{n^2}$ has radius $1$ and converges at *both* endpoints ($\sum\frac1{n^2}$ and its alternating twin). Its derived series,

$$
\sum_{n\geq1}\frac{x^{n-1}}{n} ,
$$

has the same radius $1$ — as [Theorem 11.7](#thm-b2-powerseries-calculus) guarantees — but now diverges at $x = 1$ (harmonic series) while still converging at $x = -1$ (alternating). One more differentiation yields $\sum_{n\geq2}\frac{n-1}{n}x^{n-2}$, divergent at both ends (terms do not tend to $0$). Closing insight: each differentiation multiplies the coefficients by $n$, which never moves the radius (geometric beats polynomial) but eats one order of boundary decay; term-by-term calculus is an interior sport, and whatever happens on the rim must be re-examined — the weekend problem’s Abel–Tauber theory is exactly that re-examination.

**Example 11.12 (Splitting a series by residues — worked to the end).**

Compute $f(x) = \sum_{n\geq0} \dfrac{x^{4n}}{(4n)!}$ in closed form. Both $\cosh x = \sum \frac{x^{2m}}{(2m)!}$ and $\cos x =
\sum \frac{(-1)^m x^{2m}}{(2m)!}$ have radius $\infty$, so their average may be computed term by term:

$$
\frac{\cosh x + \cos x}{2}
= \sum_{m\geq0}\frac{1 + (-1)^m}{2}\,\frac{x^{2m}}{(2m)!}
= \sum_{m \text{ even}}\frac{x^{2m}}{(2m)!}
= \sum_{n\geq0}\frac{x^{4n}}{(4n)!} = f(x) .
$$

The filter $\frac{1+(-1)^m}{2}$ keeps exactly the even $m$: this is the real avatar of the roots-of-unity filter (the complex version, with $\iu^n$, extracts residues modulo $4$ in one stroke). Closing check: $f$ solves $f'''' = f$ with $f(0) = 1$, $f'(0) = f''(0) = f'''(0) = 0$ — differentiate the series four times ([Theorem 11.7](#thm-b2-powerseries-calculus)) and watch it reproduce itself; $\frac{\cosh + \cos}{2}$ satisfies the same data.

**Definition 11.13 (Analytic functions).**

$f$ is *analytic* at $x_0$ when it is the sum of a power series in $(x - x_0)$ on a neighborhood; on an interval, when at every point. Sums of power series are analytic inside their disk (rearrangement of the expansion — admitted at this level for re-centering, the case $x_0 = 0$ being [Theorem 11.7](#thm-b2-powerseries-calculus)). Analytic implies $C^\infty$; the converse *fails*: the flat function $\eu^{-1/x^2}$ ([Exercise 11.7](#exo-b2-powerseries-7)).

**Example 11.14 (Re-centering, and the radius as a distance).**

Expand $f(x) = \frac{1}{1-x}$ around $x_0 = \frac12$: writing $x = \frac12 + h$,

$$
\frac{1}{1 - x} = \frac{1}{\frac12 - h}
= \frac{2}{1 - 2h}
= \sum_{n\geq0} 2^{n+1}\,h^n
= \sum_{n\geq0} 2^{n+1}\Bigl(x - \frac12\Bigr)^{\!n},
$$

valid for $\abs{2h} < 1$, i.e. $\abs{x - \frac12} < \frac12$. The new radius is exactly the distance from the new center to the singularity $x = 1$: re-centering shrinks (or grows) the disk to fit the nearest obstruction. Closing insight: this is the picture behind the definition of analyticity — one function, many local power series, each living on the largest disk avoiding the trouble; the Year 3 volume turns the heuristic “radius $=$ distance to the nearest complex singularity” into a theorem.

**Remark 11.15 (Common pitfalls).**

*(i) The ratio test is sufficient, not necessary:* when $\abs{a_{n+1}/a_n}$ has no limit ([Example 11.3](#ex-b2-powerseries-sinn), [Exercise 11.1](#exo-b2-powerseries-1)), go back to the definition: $R = \sup\{r : (a_nr^n)$ bounded$\}$. *(ii) Nothing crosses the boundary for free:* term-by-term differentiation and integration are theorems *inside* the [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) disk; at $\abs x = R$ each series must be re-examined (that is the whole subject of the weekend problem). *(iii) Radius of a sum:* $\min(R_a, R_b)$ is only a lower bound — cancellations can enlarge it ($a_n = 1,
b_n = -1$: sum identically $0$, radius $\infty$). *(iv) $C^\infty$ is not [analytic](#def-b2-powerseries-analytic):* a convergent Taylor series may converge to the *wrong* function ([Exercise 11.7](#exo-b2-powerseries-7)); before writing $f(x) = \sum
\frac{f^{(n)}(0)}{n!}x^n$, prove it — via an ODE ([Method 11.17](#met-b2-powerseries-ode)), a remainder estimate, or an integral formula.

**Remark 11.16 (Where this is used).**

Power series are the workhorse of three later chapters: the differential-equations chapter solves linear ODEs by injecting $\sum a_nx^n$ (the method box below, industrialized); the generating-functions chapter converts identities about probabilities into identities about power series and back; and the Year 3 volume makes the complex variable official, where analyticity becomes equivalent to complex differentiability and the “admitted re-centering” above gets its honest proof. The weekend problem explores the one place this chapter’s theorems stay silent: the boundary $\abs x = R$ itself.

**Method 11.17 (Expanding via a differential equation).**

To expand a function $f$ in a power series: find a linear ODE with polynomial coefficients satisfied by $f$; inject $\sum a_nx^n$; identify coefficients to get a recurrence for $(a_n)$; solve, and check radius and initial conditions. Example — the binomial series: $f(x) = (1+x)^\alpha$ satisfies $(1+x)f' = \alpha f$, $f(0)
= 1$; injecting gives $(n+1)a_{n+1} = (\alpha - n)a_n$, so $a_n =
\binom{\alpha}{n}$, radius $1$ (ratio test), and the sum, satisfying the same ODE with the same initial value, equals $(1 + x)^\alpha$ by the uniqueness theorem for linear ODEs (Year 1 volume).

**Example 11.18 (The method on a forced equation).**

Solve $y' = y + x$, $y(0) = 0$, by power series. Injecting $y =
\sum a_nx^n$:

$$
\sum_{n\geq0}(n+1)a_{n+1}x^n
= \sum_{n\geq0}a_nx^n + x ,
$$

and identifying coefficients: $a_1 = a_0 = 0$, $2a_2 = a_1 + 1
= 1$, and $(n+1)a_{n+1} = a_n$ for $n \geq 2$. So $a_2 =
\frac{1}{2!}$ and, by induction, $a_n = \frac{1}{n!}$ for every $n \geq 2$: radius $\infty$, and

$$
y(x) = \sum_{n\geq2}\frac{x^n}{n!} = \eu^x - 1 - x .
$$

Check: $y' = \eu^x - 1 = y + x$ and $y(0) = 0$. Closing insight: the recurrence *is* the equation, coefficient by coefficient; the forcing term only perturbs finitely many initial coefficients, after which the homogeneous pattern takes over — a discrete shadow of “particular solution plus homogeneous solution”.

## 11.3 Generating functions

**Example 11.19 (Fibonacci).**

Let $F(x) = \sum_{n\geq0} F_n x^n$ (Fibonacci numbers, $F_0 = 0$, $F_1 = 1$). The recurrence $F_{n+2} = F_{n+1} + F_n$ translates, multiplying by $x^{n+2}$ and summing, into

$$
F(x) - x = x\,F(x) + x^2 F(x)
\quad\Longrightarrow\quad
F(x) = \frac{x}{1 - x - x^2} ,
$$

valid where the series converges. The radius is $\frac{1}{\varphi}$: from $F_n \sim
\frac{\varphi^n}{\sqrt5}$ (Binet, next example — or the crude induction $F_n \leq 2^n$ plus the recurrence), the ratio test gives

$$
\frac{F_{n+1}\abs x^{n+1}}{F_n\abs x^n}
\longrightarrow \varphi\abs x ,
\qquad\text{convergence iff } \abs x < \frac1\varphi
\approx 0.618 .
$$

Partial fractions on $\frac{x}{1 - x - x^2}$ and the geometric series re-derive Binet’s formula — [generating functions](#ex-b2-powerseries-fibonacci) industrialize linear recurrences.

**Example 11.20 (Binet’s formula, carried out).**

Let $\varphi = \frac{1+\sqrt5}{2}$ and $\psi =
\frac{1-\sqrt5}{2}$, the roots of $X^2 = X + 1$; since $\varphi
+ \psi = 1$ and $\varphi\psi = -1$,

$$
1 - x - x^2 = (1 - \varphi x)(1 - \psi x) .
$$

Partial fractions: seeking $\frac{x}{(1-\varphi x)(1-\psi x)} =
\frac{A}{1 - \varphi x} + \frac{B}{1 - \psi x}$, the constant term gives $A + B = 0$ and the $x$-coefficient $-A\psi - B\varphi
= 1$, so $A(\varphi - \psi) = 1$: $A = \frac{1}{\sqrt5} = -B$. Two geometric series later,

$$
F(x) = \frac{1}{\sqrt5}\sum_{n\geq0}
\bigl(\varphi^n - \psi^n\bigr)x^n
\quad\Longrightarrow\quad
F_n = \frac{\varphi^n - \psi^n}{\sqrt5}
$$

by uniqueness of coefficients ([Theorem 11.7](#thm-b2-powerseries-calculus)). Since $\abs\psi < 1$, the term $\frac{\psi^n}{\sqrt5}$ has absolute value $< \frac12$: $F_n$ is the *nearest integer* to $\frac{\varphi^n}{\sqrt5}$. Closing insight: the radius $\frac1\varphi$ of $F$ is the reciprocal of the dominant root — growth of coefficients and [radius of convergence](#def-b2-powerseries-radius) are the same information read in opposite directions.

**Example 11.21 (Catalan numbers).**

The [Catalan numbers](#ex-b2-powerseries-catalan) $C_n$ (number of triangulations, of bracketings, of Dyck paths, …) satisfy $C_0 = 1$ and $C_{n+1} = \sum_{k=0}^n
C_kC_{n-k}$. The [generating function](#ex-b2-powerseries-fibonacci) $C(x) = \sum C_nx^n$ then satisfies ([Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini)!)

$$
C(x) = 1 + x\,C(x)^2
\quad\Longrightarrow\quad
C(x) = \frac{1 - \sqrt{1 - 4x}}{2x} ,
$$

choosing the root with $C(0) = 1$: solving the quadratic $xC^2 - C + 1 = 0$ gives the two candidates $\frac{1 \pm
\sqrt{1-4x}}{2x}$, and as $x \to 0$ the “$+$” root blows up like $\frac1x$ while the “$-$” root tends to $1$ (expand $\sqrt{1-4x} = 1 - 2x + O(x^2)$) — only the minus sign can carry a power series with $C_0 = 1$. Expanding $\sqrt{1 - 4x}$ by the binomial series gives the closed form

$$
C_n = \frac{1}{n+1}\binom{2n}{n} ,
$$

carried out in [Exercise 11.8](#exo-b2-powerseries-8).

**Remark 11.22 (Formal series versus convergent series).**

Every generating-function computation above ends by invoking the uniqueness of coefficients, and that theorem lives *inside* a disk of positive radius: before “reading off” $F_n$ or $C_n$, one must know $R > 0$. A crude a priori bound suffices — $F_n \leq 2^n$ (immediate induction) gives $R \geq
\frac12$ for Fibonacci; $C_n \leq 4^n$ (each Catalan number counts subsets of paths) gives $R \geq \frac14$. Beware the degenerate end of the scale: $\sum n!\,x^n$ has radius $0$, and manipulating it as a function is meaningless — identities involving such series belong to the *formal* calculus of coefficients, a purely algebraic game with its own (different) rules. At this level: always secure a positive radius first, then compute freely inside it.

**Remark 11.23 (Perspectives within this volume).**

Power series are one of the book’s two great expansion machines; the other is the Fourier series of the harmonic chapters, and comparing them is instructive. A power series is rigid: its coefficients are forced ($a_n = f^{(n)}(0)/n!$), its convergence is ruthless (normal inside, hopeless outside), and its sum is [analytic](#def-b2-powerseries-analytic) — infinitely rigid ([Definition 11.13](#def-b2-powerseries-analytic)). A Fourier series is supple: it represents merely piecewise-smooth signals, at the price of delicate convergence questions at the boundary of smoothness. The two theories meet in this chapter’s weekend problem: Cesàro and [Abel summation](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-abel), developed here for the boundary circle, return in the Fourier chapter as Fejér’s and Poisson’s kernels. Meanwhile the differential-equations chapter consumes power series directly ($\eu^{tA}$, series solutions), and the generating-functions chapter turns [Example 11.19](#ex-b2-powerseries-fibonacci)’s trick into a systematic calculus for probabilities.

## 11.4 Exercises

**Exercise 11.1 ★.**

Radii of convergence: $\sum \dfrac{n^2}{2^n}z^n$; $\;\sum \dfrac{z^n}{\binom{2n}{n}}$; $\;\sum z^{n!}$; $\;\sum \bigl(2 + (-1)^n\bigr)^n z^n$.

**Solution of Exercise 11.1.**

$\sum \frac{n^2}{2^n}z^n$: ratio $\frac{(n+1)^2}{2^{n+1}}\cdot
\frac{2^n}{n^2} \to \frac12$: $R = 2$.

$\sum \frac{z^n}{\binom{2n}{n}}$: $\binom{2n}{n} \sim
\frac{4^n}{\sqrt{\pi n}}$ ([Example 6.14](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#ex-b2-comparison-centralbinomial)), so $\abs{a_n}^{-1} \approx 4^n$ up to polynomial factors: $R = 4$ (ratio test: $\frac{\binom{2n}{n}}{\binom{2n+2}{n+1}} =
\frac{(n+1)^2}{(2n+1)(2n+2)} \to \frac14$).

$\sum z^{n!}$: coefficients $a_k = 1$ if $k = n!$, else $0$. For $\abs z < 1$, $\sum \abs z^{n!}$ converges (dominated by geometric); for $\abs z \geq 1$ the terms do not tend to $0$: $R = 1$.

$\sum (2 + (-1)^n)^n z^n$: coefficients $3^n$ (even $n$) and $1$ (odd $n$). Boundedness of $a_nr^n$ requires $3r \leq 1$; and $r <
\frac13$ works: $R = \frac13$.

**Exercise 11.2 ★.**

Show that $\sum z^n$, $\sum \frac{z^n}{n}$, $\sum \frac{z^n}{n^2}$ all have radius $1$ but behave differently at $z = 1$ and $z = -1$: divergence/divergence, divergence/convergence, convergence/convergence.

**Solution of Exercise 11.2.**

All three have radius $1$ (ratio test). At $z = 1$: $\sum 1$ diverges; $\sum\frac1n$ diverges; $\sum\frac{1}{n^2}$ converges. At $z = -1$: $\sum(-1)^n$ diverges; $\sum\frac{(-1)^n}{n}$ converges (alternating); $\sum\frac{(-1)^n}{n^2}$ converges ([absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def)). Boundary behavior is invisible to the radius.

**Exercise 11.3 ★.**

Compute the sums, for $\abs x < 1$:

$$
\sum_{n\geq0} n x^n,
\qquad
\sum_{n\geq0} n^2 x^n,
\qquad
\sum_{n\geq0} \frac{x^{2n+1}}{2n+1} .
$$

**Solution of Exercise 11.3.**

From $\frac{1}{1-x} = \sum x^n$, differentiate and multiply by $x$ ([Theorem 11.7](#thm-b2-powerseries-calculus)):

$$
\sum n x^n = \frac{x}{(1-x)^2} .
$$

Differentiate once more, multiply by $x$ again:

$$
\sum n^2 x^n = x\,\frac{\dd}{\dd x}\Bigl(\frac{x}{(1-x)^2}\Bigr)
= \frac{x(1 + x)}{(1-x)^3} .
$$

Third sum: it is the odd part of $-\ln(1 - x)$:

$$
\sum_{n\geq0} \frac{x^{2n+1}}{2n+1}
= \frac{-\ln(1-x) + \ln(1+x)}{2}
= \frac12 \ln\frac{1+x}{1-x}
= \operatorname{artanh} x .
$$

**Exercise 11.4 ★★.**

Expand in power series at $0$, with radius: $\dfrac{1}{(1-x)(2-x)}$ (partial fractions); $\;\ln(1 + x + x^2)$ *(write $1 + x + x^2
= \frac{1 - x^3}{1 - x}$)*.

**Solution of Exercise 11.4.**

Partial fractions: $\frac{1}{(1-x)(2-x)} = \frac{1}{1-x} -
\frac{1}{2 - x} = \sum x^n - \frac12\sum \bigl(\frac x2\bigr)^n$:

$$
\frac{1}{(1-x)(2-x)} = \sum_{n\geq0} \Bigl(1 -
\frac{1}{2^{n+1}}\Bigr)x^n,
\qquad R = 1 .
$$

$\ln(1 + x + x^2) = \ln\frac{1 - x^3}{1 - x} = \ln(1 - x^3) -
\ln(1 - x) = \sum_{n\geq1}\frac{x^n}{n} -
\sum_{m\geq1}\frac{x^{3m}}{m}$: the coefficient of $x^n$ is $\frac1n$ if $3 \nmid n$, and $\frac1n - \frac{3}{n} = -\frac2n$ if $3 \mid n$. Radius $1$ (the nearest obstruction: the series for $\ln(1-x^3)$).

**Exercise 11.5 ★★.**

Prove that $f(x) = \sum_{n\geq1} H_n x^n = -\dfrac{\ln(1 -
x)}{1 - x}$ for $\abs x < 1$, where $H_n$ is the harmonic number *([Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini) of $\sum x^n$ and $\sum \frac{x^n}{n}$)*.

**Solution of Exercise 11.5.**

[Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini) of $\sum_{m \geq 0} x^m$ (coefficients $1$) and $\sum_{k\geq1} \frac{x^k}{k}$ (coefficients $\frac1k$, $k \geq
1$), both [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) convergent for $\abs x < 1$: the coefficient of $x^n$ in the product is $\sum_{k=1}^{n} \frac1k \cdot 1 = H_n$. Hence

$$
\Bigl(\sum x^m\Bigr)\Bigl(\sum \frac{x^k}{k}\Bigr)
= \frac{1}{1-x}\cdot\bigl(-\ln(1-x)\bigr)
= \sum_{n\geq1} H_n x^n .
$$

**Exercise 11.6 ★★.**

Solve by [generating function](#ex-b2-powerseries-fibonacci) the recurrence $u_0 = 1$, $u_{n+1} =
2u_n + n$: compute $U(x) = \sum u_nx^n$ in closed form, decompose, and read off $u_n = 2^{n+1} - n - 1$.

**Solution of Exercise 11.6.**

Multiply the recurrence by $x^{n+1}$ and sum ($\abs x <
\frac12$):

$$
U(x) - 1 = 2x\,U(x) + \sum_{n\geq0} n\,x^{n+1}
= 2x\,U(x) + \frac{x^2}{(1-x)^2} ,
$$

using [Exercise 11.3](#exo-b2-powerseries-3). Hence

$$
U(x) = \frac{1}{1 - 2x}\Bigl(1 + \frac{x^2}{(1-x)^2}\Bigr)
= \frac{1 - 2x + 2x^2}{(1-2x)(1-x)^2} .
$$

Partial fractions (cover-up at $x = \frac12$ gives coefficient $2$; at the double pole $x = 1$, coefficient $-1$; the middle coefficient vanishes by evaluating at $x = 0$):

$$
U(x) = \frac{2}{1-2x} - \frac{1}{(1 - x)^2} .
$$

Expanding both:

$$
u_n = 2\cdot 2^n - (n + 1) = 2^{n+1} - n - 1 .
$$

(Check: $u_0 = 1$, $u_1 = 2u_0 + 0 = 2 = 4 - 2$.)

**Exercise 11.7 ★★.**

Let $f(x) = \eu^{-1/x^2}$ for $x \neq 0$, $f(0) = 0$. Prove that $f$ is $C^\infty$ on $\R$ with $f^{(n)}(0) = 0$ for all $n$ *(show by induction that $f^{(n)}(x) = P_n\bigl(\frac1x\bigr)
\eu^{-1/x^2}$ for polynomials $P_n$, and use the growth comparison)*. Conclude that $f$ is not [analytic](#def-b2-powerseries-analytic) at $0$: its Taylor series at $0$ converges — to the wrong function.

**Solution of Exercise 11.7.**

Induction: $f'(x) = \frac{2}{x^3}\eu^{-1/x^2}$, and if $f^{(n)}(x)
= P_n(\tfrac1x)\eu^{-1/x^2}$ then

$$
f^{(n+1)}(x) =
\Bigl(-\frac{1}{x^2}\,P_n'\Bigl(\frac1x\Bigr) +
\frac{2}{x^3}\,P_n\Bigl(\frac1x\Bigr)\Bigr)\eu^{-1/x^2} :
$$

again of the stated form. At $0$: difference quotients $\frac{f^{(n)}(h)}{h} =
\frac1h P_n(\frac1h)\eu^{-1/h^2} \to 0$ as $h \to 0$, since $Q(u)\,\eu^{-u^2} \to 0$ as $u \to \pm\infty$ for any polynomial $Q$ (exponential beats powers): by induction all $f^{(n)}(0)$ exist and vanish, and each $f^{(n)}$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) at $0$ by the same limit. So $f \in C^\infty$ with zero Taylor series at $0$; the Taylor series sums to $0 \neq f$: not [analytic](#def-b2-powerseries-analytic) at $0$.

**Exercise 11.8 ★★★.**

Complete [Example 11.21](#ex-b2-powerseries-catalan): expand $\sqrt{1 - 4x}$ with the binomial series, showing

$$
\binom{1/2}{n+1}(-4)^{n+1} = -\frac{2}{n+1}\binom{2n}{n},
$$

and deduce $C_n = \frac{1}{n+1}\binom{2n}{n}$; determine the [radius of convergence](#def-b2-powerseries-radius) of $C(x)$ and the asymptotics of $C_n$ via Stirling.

**Solution of Exercise 11.8.**

Binomial series: $\sqrt{1-4x} = \sum_{k\geq0}
\binom{1/2}{k}(-4x)^k$. For $k = n + 1 \geq 1$:

$$
\begin{align*}
\binom{1/2}{n+1}(-4)^{n+1}
&= \frac{\frac12\bigl(\frac12 - 1\bigr)\cdots\bigl(\frac12 -
n\bigr)}{(n+1)!}\,(-4)^{n+1}\\
&= \frac{(-1)^n\,1\cdot3\cdots(2n-1)}{2^{n+1}(n+1)!}\,(-4)^{n+1}
= -\frac{2}{n+1}\cdot\frac{(2n)!}{n!\,n!} ,
\end{align*}
$$

using $1\cdot3\cdots(2n-1) = \frac{(2n)!}{2^n n!}$. Hence

$$
C(x) = \frac{1 - \sqrt{1-4x}}{2x}
= \frac{1}{2x}\sum_{n\geq0}\frac{2}{n+1}\binom{2n}{n}x^{n+1}
= \sum_{n\geq0} \frac{1}{n+1}\binom{2n}{n}\,x^n :
$$

$C_n = \frac{1}{n+1}\binom{2n}{n}$. Radius: $\frac14$ (the binomial series in $4x$). Asymptotics via [Example 6.14](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#ex-b2-comparison-centralbinomial):

$$
C_n \sim \frac{4^n}{\sqrt{\pi}\; n^{3/2}} .
$$

**Exercise 11.9 ★★★.**

(Abel’s radial limit theorem, special case) Suppose $\sum a_n$ converges. Prove that $\lim_{x \to 1^-} \sum_{n} a_n x^n = \sum_n
a_n$. *([Abel summation](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-abel): with $A_n$ the partial sums and $A =
\lim A_n$, write $\sum a_nx^n = (1 - x)\sum A_n x^n$; then $\sum
a_nx^n - A = (1-x)\sum (A_n - A)x^n$, split the sum at a large $N$.)* Application: $\sum \frac{(-1)^{n-1}}{n} = \ln 2$ and $\sum
\frac{(-1)^n}{2n+1} = \frac\pi4$, re-proved from the power series.

**Solution of Exercise 11.9.**

With $A_n = \sum_{k \leq n} a_k \to A$: [Abel summation](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-abel) gives, for $0 \leq x < 1$,

$$
\sum_{n=0}^{\infty} a_n x^n = (1 - x)\sum_{n=0}^{\infty} A_n x^n
$$

(both sides converge: $(A_n)$ is bounded; and the identity follows from $a_n = A_n - A_{n-1}$ and reindexing). Since $(1 -
x)\sum x^n = 1$:

$$
\sum_n a_nx^n - A = (1-x)\sum_{n} (A_n - A)x^n .
$$

Given $\varepsilon$, choose $N$ with $\abs{A_n - A} \leq
\varepsilon$ for $n > N$; then

$$
\Bigl|\sum a_nx^n - A\Bigr|
\leq (1-x)\sum_{n \leq N}\abs{A_n - A} + \varepsilon(1 -
x)\sum_{n > N}x^n
\leq (1-x)\,C_N + \varepsilon ,
$$

and letting $x \to 1^-$: limsup $\leq \varepsilon$ for all $\varepsilon$. Hence the radial limit is $A$.

Applications: $\sum \frac{(-1)^{n-1}}{n}$ converges (alternating), and for $x < 1$ its power series sums to $\ln(1 + x)$: by Abel, the sum is $\ln 2$. Likewise $\sum\frac{(-1)^n}{2n+1}x^{2n+1} =
\arctan x$ gives $\frac\pi4$ at $x = 1$ — the Year 1 integral proofs, now structural.

**Exercise 11.10 ★.**

Show that $\displaystyle\sum_{n\geq1}\frac{x^n}{n(n+1)} = 1 +
\frac{1-x}{x}\,\ln(1-x)$ for $0 < \abs x < 1$, determine the radius, and check that the convergence is normal on $\intcc{-1}{1}$; verify that the value at $x = 1$ predicted by [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) agrees with the telescoping sum $\sum
\frac{1}{n(n+1)} = 1$.

**Solution of Exercise 11.10.**

Both $\sum\frac{x^n}{n}$ and $\sum\frac{x^n}{n+1}$ have radius $1$, and $\frac{1}{n(n+1)} = \frac1n - \frac1{n+1}$, so for $0
< \abs x < 1$:

$$
\sum_{n\geq1}\frac{x^n}{n(n+1)}
= -\ln(1-x) - \frac1x\sum_{n\geq1}\frac{x^{n+1}}{n+1}
= -\ln(1-x) - \frac{-\ln(1-x) - x}{x}
= 1 + \frac{1-x}{x}\ln(1-x) .
$$

Radius $1$; and $\norm{x^n/(n(n+1))}_{\infty,\intcc{-1}1} =
\frac{1}{n(n+1)}$ is [summable](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-summable): [normal convergence](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) on $\intcc{-1}{1}$, so the sum is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) there. As $x \to
1^-$, $(1-x)\ln(1-x) \to 0$ and the closed form tends to $1$ — in agreement with the telescoping value $\sum\frac{1}{n(n+1)} =
\lim_N\bigl(1 - \frac{1}{N+1}\bigr) = 1$ at $x = 1$.

**Exercise 11.11 ★★.**

(Derangements) Let $D_n$ be the number of permutations of $n$ objects without fixed point ($D_0 = 1$). Sorting the permutations of $\{1, \dots, n\}$ by their fixed-point set gives $n! = \sum_{k=0}^{n}\binom nk D_{n-k}$. Multiply by $\frac{x^n}{n!}$, sum, and recognize a [Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini) to obtain the exponential [generating function](#ex-b2-powerseries-fibonacci)

$$
\sum_{n\geq0} D_n\,\frac{x^n}{n!} = \frac{\eu^{-x}}{1-x}
\qquad (\abs x < 1),
$$

then read off the closed form $\dfrac{D_n}{n!} =
\sum_{k=0}^{n}\dfrac{(-1)^k}{k!}$ and the limit $\dfrac{D_n}{n!}
\to \eu^{-1}$.

**Solution of Exercise 11.11.**

Sorting the $n!$ permutations by their fixed-point set: choosing the $k$ fixed points ($\binom nk$ ways) and deranging the other $n - k$ objects gives $n! = \sum_{k=0}^n\binom nk D_{n-k}$. Dividing by $n!$:

$$
1 = \sum_{k=0}^{n}\frac{1}{k!}\cdot\frac{D_{n-k}}{(n-k)!} ,
$$

which says exactly that the [Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini) of $\eu^x =
\sum\frac{x^k}{k!}$ and $D(x) = \sum D_n\frac{x^n}{n!}$ is $\sum x^n = \frac{1}{1-x}$. Both factors converge [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) for $\abs x < 1$ ($D_n \leq n!$, so $D$ is dominated by the geometric series): the product identity is legitimate ([Proposition 11.4](#prop-b2-powerseries-operations)), and

$$
D(x) = \frac{\eu^{-x}}{1-x} .
$$

[Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini) of $\eu^{-x} = \sum\frac{(-1)^kx^k}{k!}$ and $\sum x^m$: the coefficient of $x^n$ is $\sum_{k=0}^{n}\frac{(-1)^k}{k!}$, and by uniqueness of power series coefficients ([Theorem 11.7](#thm-b2-powerseries-calculus)):

$$
\frac{D_n}{n!} = \sum_{k=0}^{n}\frac{(-1)^k}{k!}
\xrightarrow[n\to\infty]{} \eu^{-1} :
$$

about $37\%$ of all permutations are derangements, whatever $n$.

**Exercise 11.12 ★★★.**

Prove, with the binomial series of [Method 11.17](#met-b2-powerseries-ode), that

$$
\frac{1}{\sqrt{1 - 4x}} = \sum_{n\geq0}\binom{2n}{n}x^n
\qquad \Bigl(\abs x < \frac14\Bigr),
$$

and deduce, by squaring ([Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini) against $\frac{1}{1-4x} = \sum 4^nx^n$), the convolution identity

$$
\sum_{k=0}^{n}\binom{2k}{k}\binom{2n-2k}{n-k} = 4^n .
$$

**Solution of Exercise 11.12.**

Binomial series with $\alpha = -\frac12$ at $-4x$:

$$
\binom{-1/2}{n}(-4)^n
= \frac{\bigl(-\frac12\bigr)\bigl(-\frac32\bigr)\cdots
\bigl(-\frac{2n-1}2\bigr)}{n!}(-4)^n
= \frac{1\cdot3\cdots(2n-1)}{2^n\,n!}\,4^n
= \frac{(2n)!}{2^n n!}\cdot\frac{2^n}{n!}
= \binom{2n}{n},
$$

using $1\cdot3\cdots(2n-1) = \frac{(2n)!}{2^nn!}$. Hence $(1-4x)^{-1/2} = \sum\binom{2n}nx^n$ for $\abs{4x} < 1$. Squaring ([Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini), legitimate by absolute convergence) and comparing with $\frac{1}{1-4x} = \sum 4^nx^n$: the coefficient of $x^n$ in the square is $\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}$, and uniqueness of coefficients gives

$$
\sum_{k=0}^{n}\binom{2k}{k}\binom{2n-2k}{n-k} = 4^n .
$$

## 11.5 Problem: Abel, Tauber, and the boundary of convergence

**Problem 11.1.**

Inside the disk of convergence everything is easy; all the drama of power series happens *on* the boundary. This problem builds the boundary theory in the real variable: Abel’s theorem in its uniform form, its converse under Tauber’s condition, the Cesàro–Abel hierarchy of summation methods (with Frobenius’ theorem), term-by-term integration up to the boundary with classical constants as dividends, and finally the rigidity of [analytic functions](#def-b2-powerseries-analytic) — the identity theorem. Throughout, $(a_n)$ is a real sequence, $f(x) = \sum_{n\geq0}
a_nx^n$, and $A_n = a_0 + \dots + a_n$.

**Part I — Abel’s theorem, [uniformly](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def).** Assume in this part that $\sum a_n$ converges, and set $r_n =
\sum_{k\geq n} a_k$ (so $r_n \to 0$ and $a_n = r_n -
r_{n+1}$).

1. Prove, by summation by parts, that for all $0 \leq x  \leq 1$ and $N \leq M$: $$\Bigl|\sum_{n=N}^{M} a_n x^n\Bigr|  \leq 2\sup_{n \geq N}\,\abs{r_n} .$$
2. Deduce that $\sum a_nx^n$ converges *[uniformly](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def)* on $\intcc{0}{1}$ , that its sum is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) there, and recover the radial limit of [Exercise 11.9](#exo-b2-powerseries-9) : $f(x) \to \sum a_n$ as $x  \to 1^-$ .
3. (Abel’s theorem for [Cauchy products](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini) ) Let $\sum a_n = A$ , $\sum b_n = B$ and suppose the [Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini) $\sum  c_n$ , $c_n = \sum_{k} a_kb_{n-k}$ , *converges* , with sum $C$ . Prove $C = AB$ *(inside the disk the product identity holds by [Proposition 11.4](#prop-b2-powerseries-operations); let $x \to  1^-$)* .
4. Show the hypothesis matters: for $a_n = b_n =  \frac{(-1)^n}{\sqrt{n+1}}$ , both series converge, yet $\abs{c_n} \geq \frac{2(n+1)}{n+2} \geq 1$ *(bound each factor $\sqrt{(k+1)(n-k+1)}$ by AM–GM)* : the [Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini) of two convergent series may diverge.
5. (A dividend of [Exercise 11.5](#exo-b2-powerseries-5)) Show that $\bigl(\ln(1-x)\bigr)^2 = 2\sum_{n\geq1}  \frac{H_n}{n+1}x^{n+1}$ on $\intoo{-1}{1}$, check that $\bigl(\frac{H_n}{n+1}\bigr)_{n\geq1}$ decreases to $0$, and conclude with Abel: $$\sum_{n\geq1} (-1)^{n+1}\,\frac{H_n}{n+1}  = \frac{(\ln 2)^2}{2} .$$

**Part II — Tauber’s converse.** Call $\sum
a_n$ *Abel-summable* to $L$ when $f(x) \to L$ as $x \to
1^-$.

6. Show that $\sum (-1)^n$ is Abel-summable to $\frac12$ yet divergent: Abel’s theorem has no unconditional converse.
7. (Cesàro lemma) If $u_n \to 0$ then $\frac{u_1 + \dots  + u_N}{N} \to 0$ *(split the sum at a fixed $m$)* .
8. Assume now $n\,a_n \to 0$ and $f(x) \to L$. With $x_N =  1 - \frac1N$, prove the two estimates $$\Bigl|\sum_{n=0}^{N} a_n\bigl(1 - x_N^n\bigr)\Bigr|  \leq \frac{1}{N}\sum_{n=1}^{N} n\,\abs{a_n},  \qquad  \Bigl|\sum_{n>N} a_n x_N^n\Bigr|  \leq \sup_{n>N}\bigl(n\abs{a_n}\bigr)$$ *(for the first, $1 - x^n \leq n(1-x)$; for the second, $\abs{a_n} \leq \frac{1}{N}\sup_{m>N}  m\abs{a_m}$ and $\sum x_N^n \leq N$)*.
9. Conclude *Tauber’s theorem* : if $n\,a_n \to 0$ and $\sum a_n$ is Abel-summable to $L$ , then $\sum a_n$ converges to $L$ .
10. (The easy Tauberian for positive coefficients) If $a_n  \geq 0$ and $f$ is bounded on $\intco{0}{1}$ , show that $\sum a_n$ converges and $\sum a_n = \lim_{x\to1^-}  f(x)$ *(bound $\sum_{n\leq N}a_nx^n \leq f(x)$ and let $x \to 1^-$, then use Abel)* .

**Part III — Cesàro means and Frobenius’ theorem.** Call $\sum a_n$ *Cesàro-summable* to $L$ when $\sigma_N = \frac{A_0 + \dots + A_{N-1}}{N} \to L$.

11. Show that a convergent series is Cesàro-summable to its sum *(question 7 applied to $A_n - L$)* .
12. Compute the Cesàro value of $\sum(-1)^n$ and check it agrees with the Abel value $\frac12$ of question 6.
13. With $S_n = A_0 + \dots + A_n = (n+1)\,\sigma_{n+1}$, prove the two identities, for $0 \leq x < 1$: $$f(x) = (1-x)^2\sum_{n\geq0}(n+1)\,\sigma_{n+1}x^n,  \qquad  (1-x)^2\sum_{n\geq0}(n+1)x^n = 1 .$$
14. (Frobenius) Deduce: if $\sigma_N \to L$ then $f(x) \to  L$ as $x \to 1^-$ — Cesàro-summable implies Abel-summable, to the same value *(subtract the two identities and split the sum at a large $N$, as in [Exercise 11.9](#exo-b2-powerseries-9))* .
15. Show that the hierarchy $$\text{convergent} \;\Longrightarrow\;  \text{Ces\`aro-summable} \;\Longrightarrow\;  \text{Abel-summable}$$ is strict at both arrows: question 6 for the first; for the second, show that $\sum(-1)^n(n+1)$ is Abel-summable to $\frac14$ (compute $f$) but not Cesàro-summable (compute $\sigma_N$ separately for even and odd $N$).

**Part IV — Integrating to the boundary.**

16. Suppose $\sum a_nx^n$ converges on $\intco{0}{1}$ and $\sum \frac{a_n}{n+1}$ converges. Prove that the [improper integral](https://one-course.com/books/math/4/en/chapter/9-integration#def-b2-integration-improper) $\int_0^1 f$ exists and $$\int_0^1 \Bigl(\sum_{n\geq0} a_nx^n\Bigr)\dd x  = \sum_{n\geq0}\frac{a_n}{n+1}$$ *(the primitive $F(x) = \sum\frac{a_n}{n+1}x^{n+1}$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) at $1$ by Part I)*.
17. Let $\eta = \sum_{n\geq1}\frac{(-1)^{n-1}}{n^2}$ . Show $\int_0^1 \frac{\ln(1+x)}{x}\dd x = \eta$ and, by splitting even and odd indices in the [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) convergent $\sum \frac1{n^2}$ , that $\eta =  \frac12\sum_{n\geq1}\frac{1}{n^2}$ . (The Fourier chapter’s weekend problem evaluates $\sum\frac1{n^2} = \frac{\pi^2}{6}$ .)
18. Prove $$\sum_{n\geq0}\frac{(-1)^n}{3n+1}  = \int_0^1\frac{\dd x}{1+x^3}  = \frac13\Bigl(\ln 2 + \frac{\pi}{\sqrt3}\Bigr)$$ *(the series converges by Leibniz; integrate the geometric series $\sum(-1)^nx^{3n}$ with question 16; then partial fractions: $\frac{1}{1+x^3} =  \frac{1/3}{1+x} + \frac{(2-x)/3}{x^2-x+1}$)*.
19. From the binomial series for $(1-t)^{-1/2}$ ([Exercise 11.12](#exo-b2-powerseries-12)) derive $$\arcsin x = \sum_{n\geq0}  \frac{\binom{2n}{n}}{4^n(2n+1)}\,x^{2n+1}  \quad(\abs x < 1),  \qquad\text{then}\qquad  \sum_{n\geq0}\frac{\binom{2n}{n}}{4^n(2n+1)}  = \frac\pi2 ,$$ justifying the boundary value by *normal* convergence on $\intcc{-1}{1}$ (use $\binom{2n}n4^{-n} \sim \frac{1}{\sqrt{\pi n}}$, [Example 6.14](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#ex-b2-comparison-centralbinomial)) — here not even Abel is needed.
20. (Catalan at the boundary) Show that $\sum C_n 4^{-n} =  2$ : the Catalan series of [Example 11.21](#ex-b2-powerseries-catalan) converges *at* its radius $\frac14$ (asymptotics of [Exercise 11.8](#exo-b2-powerseries-8) ), its sum is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $\intcc{0}{\frac14}$ , and the closed form has limit $2$ there.

**Part V — Rigidity: the identity theorem.**

21. (Isolated zeros) Let $f = \sum a_nx^n$ have radius $R >  0$ and not all $a_n = 0$ ; let $m$ be the smallest index with $a_m \neq 0$ . Show $f(x) = x^m g(x)$ with $g$ a power series of radius $R$ , $g(0) = a_m \neq 0$ , and deduce that $f$ has no zero in some punctured neighborhood of $0$ .
22. (Identity theorem) Let $f, h$ be sums of power series near $0$ and $(x_k)$ a sequence of *nonzero* points with $x_k \to 0$ and $f(x_k) = h(x_k)$ . Prove that $f$ and $h$ have the same coefficients, hence coincide near $0$ .
23. Find *all* functions $f$ [analytic](#def-b2-powerseries-analytic) near $0$ with $$f\Bigl(\frac1k\Bigr) = \frac{k^2}{k^2+1}  \qquad\text{for all large integers } k .$$
24. Show that a function [analytic](#def-b2-powerseries-analytic) on an [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) interval $I$ which vanishes on a subinterval vanishes identically on $I$ *(the set of points around which $f$ vanishes identically is [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) and, by the identity theorem applied at accumulation points, closed in $I$)* . Conclude that no nonzero [analytic function](#def-b2-powerseries-analytic) on $\R$ has [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) support — whereas $C^\infty$ bump functions exist ( [Exercise 11.7](#exo-b2-powerseries-7) supplies the building block): analyticity is rigid, smoothness is flabby.
25. Synthesis. In one sentence each: (i) what Abel’s theorem adds to the normal-convergence package of [Lemma 11.1](#lem-b2-powerseries-abel) ; (ii) the exact hypotheses under which the converse holds (Tauber) and the intermediate rung (Frobenius); (iii) one boundary constant from Part IV you could now derive for a friend in two lines; (iv) where Cesàro means will reappear in this book, for series of a very different kind.

**Solution of Problem 11.1.**

**1.** With $a_n = r_n - r_{n+1}$, summation by parts:

$$
\sum_{n=N}^{M} a_nx^n
= r_Nx^N + \sum_{n=N+1}^{M} r_n\bigl(x^n - x^{n-1}\bigr)
- r_{M+1}x^M .
$$

For $0 \leq x \leq 1$ the increments $x^{n-1} - x^n$ are nonnegative and telescope to $x^N - x^M$; with $s = \sup_{n\geq
N}\abs{r_n}$:

$$
\Bigl|\sum_{n=N}^{M}a_nx^n\Bigr|
\leq s\bigl(x^N + (x^N - x^M) + x^M\bigr) = 2s\,x^N \leq 2s .
$$

**2.** Since $r_n \to 0$, $\sup_{n\geq N}\abs{r_n} \to 0$: question 1 is exactly the uniform Cauchy criterion on $\intcc{0}{1}$, so $\sum a_nx^n$ converges [uniformly](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) there and its sum is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) ([Theorem 10.11](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#thm-b2-funcseq-seriestransfer)). The value at $1$ being $\sum a_n$, [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) at $1$ is the radial limit of [Exercise 11.9](#exo-b2-powerseries-9).

**3.** For $\abs x < 1$ all three power series converge [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) and $\bigl(\sum a_nx^n\bigr)\bigl(\sum b_nx^n\bigr)
= \sum c_nx^n$ ([Proposition 11.4](#prop-b2-powerseries-operations)). By question 2 each factor and the product side are [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $\intcc{0}{1}$ (their coefficient series converge by hypothesis); letting $x \to 1^-$ in the identity: $AB = C$.

**4.** Here

$$
\abs{c_n} = \sum_{k=0}^{n}
\frac{1}{\sqrt{(k+1)(n-k+1)}}
\geq \sum_{k=0}^{n}\frac{2}{n+2}
= \frac{2(n+1)}{n+2} \geq 1,
$$

by AM–GM: $\sqrt{(k+1)(n-k+1)} \leq \frac{(k+1) +
(n-k+1)}{2} = \frac{n+2}{2}$. The general term of $\sum c_n$ does not tend to $0$: the [Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini) diverges, although both factors converge (alternating series).

**5.** [Exercise 11.5](#exo-b2-powerseries-5) gives $\frac{-\ln(1-x)}{1-x} = \sum H_nx^n$ ($\abs x < 1$). Term-by-term primitives ([Theorem 11.7](#thm-b2-powerseries-calculus) (2)), both sides vanishing at $0$:

$$
\frac{\bigl(\ln(1-x)\bigr)^2}{2}
= \sum_{n\geq1}\frac{H_n}{n+1}\,x^{n+1} .
$$

Decrease: $(n+2)H_n \geq (n+1)H_{n+1}$ amounts to $H_n \geq
1$, true for $n \geq 1$; and $\frac{H_n}{n+1} \sim \frac{\ln
n}{n} \to 0$: at $x = -1$ the series converges by the alternating test. Substituting $x \mapsto -x$ and applying question 2 at $x = 1$:

$$
\frac{(\ln 2)^2}{2}
= \sum_{n\geq1}\frac{H_n}{n+1}(-1)^{n+1},
$$

the announced value.

**6.** $f(x) = \sum(-1)^nx^n = \frac{1}{1+x} \to \frac12$ as $x \to 1^-$: Abel-summable to $\frac12$. But the partial sums are $1, 0, 1, 0, \dots$: divergent.

**7.** Given $\varepsilon > 0$ choose $m$ with $\abs{u_n}
\leq \varepsilon$ for $n > m$; for $N \geq m$:

$$
\Bigl|\frac{u_1 + \dots + u_N}{N}\Bigr|
\leq \frac{\abs{u_1} + \dots + \abs{u_m}}{N} +
\varepsilon\,\frac{N - m}{N}
\leq \frac{C_m}{N} + \varepsilon,
$$

so $\limsup \leq \varepsilon$ for every $\varepsilon$: the means tend to $0$.

**8.** For $0 \leq x \leq 1$: $1 - x^n = (1-x)(1 + x +
\dots + x^{n-1}) \leq n(1-x)$, so

$$
\Bigl|\sum_{n=0}^N a_n(1 - x_N^n)\Bigr|
\leq (1 - x_N)\sum_{n=1}^N n\abs{a_n}
= \frac1N\sum_{n=1}^{N}n\abs{a_n} .
$$

For $n > N$: $\abs{a_n} = \frac{n\abs{a_n}}{n} \leq
\frac{1}{N}\sup_{m>N}m\abs{a_m}$, and $\sum_{n>N}x_N^n \leq
\frac{1}{1 - x_N} = N$:

$$
\Bigl|\sum_{n>N}a_nx_N^n\Bigr|
\leq \frac{\sup_{m>N}m\abs{a_m}}{N}\cdot N
= \sup_{m>N}\,m\abs{a_m} .
$$

**9.** Decompose

$$
A_N - L = \sum_{n=0}^{N}a_n\bigl(1 - x_N^n\bigr)
- \sum_{n>N}a_nx_N^n + \bigl(f(x_N) - L\bigr) .
$$

The first term tends to $0$ by question 7 (the means of $n\abs{a_n} \to 0$), the second by question 8 (the sup tends to $0$), the third because $x_N \to 1^-$ and $f(x) \to L$. Hence $A_N \to L$: Tauber’s theorem.

**10.** For $x \in \intco{0}{1}$ and any $N$: $\sum_{n\leq N}a_nx^n \leq f(x) \leq M$ (nonnegative terms). Let $x \to 1^-$ in the finite sum: $\sum_{n\leq N}a_n \leq M$. The partial sums are increasing and bounded: $\sum a_n$ converges, and then question 2 gives $\lim_{x\to1^-}f(x) =
\sum a_n$.

**11.** $\sigma_N - L$ is the mean of the $N$ numbers $A_n - L$ ($0 \leq n < N$), which tend to $0$: question 7.

**12.** $A_n = 1$ for even $n$, $0$ for odd: $A_0 + \dots
+ A_{N-1} = \lceil N/2\rceil$, so $\sigma_N =
\frac{\lceil N/2\rceil}{N} \to \frac12$, the Abel value of question 6.

**13.** Under $\sigma_N \to L$ one has $S_n = O(n)$, hence $A_n = S_n - S_{n-1} = O(n)$ and $a_n = O(n)$: all the series below have radius $\geq 1$. For $\abs x < 1$, from $a_n = A_n -
A_{n-1}$ and $A_nx^n \to 0$:

$$
(1-x)\sum_n A_nx^n = \sum_n A_nx^n - \sum_n A_nx^{n+1}
= \sum_n a_nx^n = f(x),
$$

and identically $(1-x)\sum S_nx^n = \sum A_nx^n$, so $f(x) =
(1-x)^2\sum_n S_nx^n = (1-x)^2\sum_n(n+1)\sigma_{n+1}x^n$. Finally $\sum(n+1)x^n = \frac{1}{(1-x)^2}$ ([Exercise 11.3](#exo-b2-powerseries-3)), which is the second identity.

**14.** Subtracting $L$ times the second identity from the first:

$$
f(x) - L = (1-x)^2\sum_{n\geq0}(n+1)
\bigl(\sigma_{n+1} - L\bigr)x^n .
$$

Given $\varepsilon$, choose $N$ with $\abs{\sigma_{n+1} - L}
\leq \varepsilon$ for $n \geq N$; then

$$
\abs{f(x) - L} \leq (1-x)^2 C_N +
\varepsilon(1-x)^2\sum_{n}(n+1)x^n
= (1-x)^2C_N + \varepsilon ,
$$

and letting $x \to 1^-$: $\limsup \leq \varepsilon$. Hence $f(x) \to L$: Frobenius’ theorem.

**15.** $f(x) = \sum(-1)^n(n+1)x^n = \frac{1}{(1+x)^2}$ (differentiate the geometric series at $-x$): Abel value $\frac14$. Partial sums: $A_{2k} = k+1$, $A_{2k+1} = -(k+1)$ (immediate induction). Then $S_{2m-1} = 0$ (consecutive pairs cancel) and $S_{2m} = m + 1$, so

$$
\sigma_{2m} = \frac{S_{2m-1}}{2m} = 0,
\qquad
\sigma_{2m+1} = \frac{m+1}{2m+1} \to \frac12 :
$$

$(\sigma_N)$ has two distinct cluster values: not Cesàro-summable. With questions 6 and 11–14, the hierarchy convergent $\Rightarrow$ Cesàro $\Rightarrow$ Abel is strict at both arrows.

**16.** The primitive series $F(x) =
\sum\frac{a_n}{n+1}x^{n+1}$ has the same radius and $F' = f$ on $\intco{0}{1}$ ([Theorem 11.7](#thm-b2-powerseries-calculus)); since $\sum\frac{a_n}{n+1}$ converges, Part I (question 2) makes $F$ [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $\intcc{0}{1}$. As $\int_0^x f = F(x)$ (equal derivatives, equal value $0$ at $0$),

$$
\int_0^x f \xrightarrow[x\to1^-]{} F(1)
= \sum_{n\geq0}\frac{a_n}{n+1} :
$$

the [improper integral](https://one-course.com/books/math/4/en/chapter/9-integration#def-b2-integration-improper) exists with the stated value.

**17.** $\frac{\ln(1+x)}{x} =
\sum_{n\geq1}\frac{(-1)^{n-1}}{n}x^{n-1}$ (radius $1$; [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) at $0$). The series of $\frac{a_m}{m+1}$ is $\sum_{n\geq1}\frac{(-1)^{n-1}}{n^2}$, [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) convergent: question 16 gives $\int_0^1\frac{\ln(1+x)}{x}\dd x = \eta$. In the [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) convergent $\sum\frac1{n^2}$, regroup even and odd:

$$
\eta = \sum_{\text{odd}}\frac1{n^2} -
\sum_{\text{even}}\frac1{n^2}
= \sum_{n}\frac1{n^2} - 2\sum_{k}\frac1{(2k)^2}
= \Bigl(1 - \frac12\Bigr)\sum_n\frac1{n^2}
= \frac12\sum_{n\geq1}\frac1{n^2} .
$$

**18.** Leibniz: $\frac{1}{3n+1}\downarrow0$, the series converges. On $\intco{0}{1}$, $\sum(-1)^nx^{3n} =
\frac{1}{1+x^3}$, and $\sum\frac{(-1)^n}{3n+1}$ converges: question 16 gives $\sum\frac{(-1)^n}{3n+1} =
\int_0^1\frac{\dd x}{1+x^3}$. Partial fractions (check: $\frac13(x^2-x+1) + \frac{2-x}{3}(1+x) = 1$):

$$
\int_0^1\frac{\dd x}{1+x^3}
= \frac13\ln2 + \frac13\int_0^1\frac{2-x}{x^2-x+1}\dd x .
$$

Writing $2 - x = -\frac12(2x-1) + \frac32$: the $\ln(x^2-x+1)$ part vanishes at both ends, and

$$
\frac32\int_0^1\frac{\dd x}{(x-\frac12)^2 + \frac34}
= \frac32\cdot\frac{2}{\sqrt3}
\Bigl[\arctan\frac{2x-1}{\sqrt3}\Bigr]_0^1
= \sqrt3\cdot\frac{\pi}{3} = \frac{\pi}{\sqrt3} .
$$

Total: $\frac13\bigl(\ln2 + \frac{\pi}{\sqrt3}\bigr)$.

**19.** Substituting $t = x^2$ in the series of [Exercise 11.12](#exo-b2-powerseries-12) and integrating term by term (the primitive of $(1-x^2)^{-1/2}$ vanishing at $0$ is $\arcsin$):

$$
\arcsin x = \sum_{n\geq0}
\frac{\binom{2n}n}{4^n(2n+1)}x^{2n+1}
\qquad(\abs x < 1) .
$$

The coefficients are $\sim \frac{1}{2\sqrt\pi\,n^{3/2}}$ ([Example 6.14](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#ex-b2-comparison-centralbinomial)), [summable](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-summable): the series converges *[normally](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series)* on $\intcc{-1}{1}$, its sum is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) there, and coincides with the [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $\arcsin$ on $\intoo{-1}{1}$, hence at $x = 1$ too:

$$
\sum_{n\geq0}\frac{\binom{2n}n}{4^n(2n+1)}
= \arcsin 1 = \frac\pi2 .
$$

**20.** $C_n4^{-n} \sim \frac{1}{\sqrt\pi\,n^{3/2}}$ ([Exercise 11.8](#exo-b2-powerseries-8)): [normal convergence](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) of $\sum
C_nx^n$ on $\intcc{0}{\frac14}$, so its sum is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) there; on $\intoo{0}{\frac14}$ it equals $\frac{1-\sqrt{1-4x}}{2x}$ ([Example 11.21](#ex-b2-powerseries-catalan)), whose limit at $\frac14^-$ is $\frac{1-0}{1/2} = 2$. Hence $\sum_{n\geq0}
C_n4^{-n} = 2$.

**21.** $f(x) = \sum_{n\geq m}a_nx^n = x^m g(x)$ with $g(x) = \sum_{k\geq0}a_{m+k}x^k$; if $(a_nr^n)$ is bounded then $(a_{m+k}r^k)$ is too (divide by $r^m$): $g$ has radius $\geq
R$. $g$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) with $g(0) = a_m \neq 0$, so $g \neq 0$ on some $\intcc{-\delta}{\delta}$, and $f(x) = x^mg(x) \neq 0$ for $0 < \abs x \leq \delta$.

**22.** $d = f - h$ is the sum of a power series near $0$ vanishing at the nonzero points $x_k \to 0$. If some coefficient of $d$ were nonzero, question 21 would give a punctured neighborhood of $0$ free of zeros of $d$ — contradicting $d(x_k) = 0$. So all coefficients of $d$ vanish: $f$ and $h$ have equal coefficients and coincide near $0$.

**23.** The function $h(x) = \frac{1}{1+x^2} =
\sum(-1)^nx^{2n}$ (radius $1$) satisfies $h(\frac1k) =
\frac{1}{1 + 1/k^2} = \frac{k^2}{k^2+1}$. Any [analytic](#def-b2-powerseries-analytic) $f$ with the same values agrees with $h$ at the points $\frac1k \to 0$: by the identity theorem (question 22), $f = \frac{1}{1+x^2}$ near $0$ — the unique solution.

**24.** Let $Z$ be the set of points of $I$ having a neighborhood on which $f$ vanishes identically: [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) by definition, nonempty (the subinterval). Closed in $I$: if $y
\in I$ is a limit of points of $Z$, then $y$ is an accumulation point of zeros of $f$; expanding $f$ in a power series at $y$ (analyticity) and applying questions 21–22 re-centered at $y$, all coefficients at $y$ vanish, so $f \equiv 0$ near $y$: $y
\in Z$. An interval is [connected](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-connected), so $Z = I$: $f \equiv 0$ on $I$. In particular an [analytic function](#def-b2-powerseries-analytic) on $\R$ vanishing outside a [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) set vanishes on an interval, hence everywhere: no nonzero [analytic](#def-b2-powerseries-analytic) bump. The $C^\infty$ world is different: gluing the flat function of [Exercise 11.7](#exo-b2-powerseries-7) (e.g. $x \mapsto
\eu^{-1/x^2}\mathbf 1_{x>0}$ and its mirror) produces smooth bumps with [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) support.

**25.** (i) [Normal convergence](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) lives on [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) subdisks strictly inside the disk; Abel’s theorem extends [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) to a boundary point, under the sole hypothesis that the coefficient series converges there. (ii) The converse holds under Tauber’s condition $na_n \to 0$ (question 9), and Cesàro [summability](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-summable) sits strictly between convergence and Abel [summability](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-summable) (Frobenius, questions 14–15). (iii) For a friend: $\sum\frac{(-1)^n}{3n+1} = \int_0^1\frac{\dd x}{1+x^3}$ by integrating the geometric series to the boundary, then partial fractions. (iv) Cesàro means return in the Fourier chapter as Fejér’s theorem, where averaging the partial sums repairs the failure of [pointwise convergence](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) — same medicine, new patient.