University Mathematics — Year 2 · Bachelor Year 2

11Power Series

Power series are the best-behaved series of functions in mathematics: inside their disk of convergence they converge normally on compacts, may be differentiated and integrated term by term without a second thought, and their sums — analytic functions — are determined by their coefficients. This chapter proves that whole package and reclaims, honestly, every Taylor series of the Year 1 volume; generating functions 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 for every $\abs z < \abs {z_0}$ , and normally 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: absolute convergence for $\abs z < R$ (normal on compact subdisks), divergence — indeed unbounded terms — for $\abs z > R$ . On the boundary circle, anything can happen (Exercise 11.2). 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).

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, legitimate by absolute convergence and Theorem 7.14.)

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 there (Lemma 11.1), so the doubly indexed family $(a_kz^k\,b_lz^l)_{k,l}$ is summable, and Theorem 7.14 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 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 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 $\frac{1}{1-x}$ sums the coefficients)

A Cauchy product 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, 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 $x \in \intoo{-R}{R}$ ).

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).
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 ( $n\abs{a_n}r^{n-1}$ summable by the radius computation); the original converges at $x = 0$ : the differentiation theorem for series (Theorem 10.11) 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 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 (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 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) 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). Analytic implies $C^\infty$ ; the converse fails: the flat function $\eu^{-1/x^2}$ (Exercise 11.7).

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, Exercise 11.1), 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 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: a convergent Taylor series may converge to the wrong function (Exercise 11.7); before writing $f(x) = \sum \frac{f^{(n)}(0)}{n!}x^n$ , prove it — via an ODE (Method 11.17), 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 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). 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 are the same information read in opposite directions.

Example 11.21 (Catalan numbers)

The Catalan numbers $C_n$ (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 $C(x) = \sum C_nx^n$ then satisfies (Cauchy product!)

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.

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 — infinitely rigid (Definition 11.13). 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, 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’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

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), 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

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). 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

Solution of Exercise 11.3.

From $\frac{1}{1-x} = \sum x^n$ , differentiate and multiply by $x$ (Theorem 11.7):

\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

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 of $\sum x^n$ and $\sum \frac{x^n}{n}$ ).

Solution

Solution of Exercise 11.5.

Cauchy product of $\sum_{m \geq 0} x^m$ (coefficients $1$ ) and $\sum_{k\geq1} \frac{x^k}{k}$ (coefficients $\frac1k$ , $k \geq 1$ ), both absolutely 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 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

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. 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 at $0$ : its Taylor series at $0$ converges — to the wrong function.

Solution

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 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 at $0$ .

Exercise 11.8 ★★★

Complete Example 11.21: 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 of $C(x)$ and the asymptotics of $C_n$ via Stirling.

Solution

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:

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: 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

Solution of Exercise 11.9.

With $A_n = \sum_{k \leq n} a_k \to A$ : Abel summation 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 agrees with the telescoping sum $\sum \frac{1}{n(n+1)} = 1$ .

Solution

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: normal convergence on $\intcc{-1}{1}$ , so the sum is continuous 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 to obtain the exponential generating function

\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

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 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 for $\abs x < 1$ ( $D_n \leq n!$ , so $D$ is dominated by the geometric series): the product identity is legitimate (Proposition 11.4), and

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

Cauchy product 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):

\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, 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 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

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, 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 — 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. 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 $a_n = r_n - r_{n+1}$ ).

Prove, by summation by parts, that for all $0 \leq x \leq 1$ and $N \leq M$ :

$\Bigl|\sum_{n=N}^{M} a_n x^n\Bigr| \leq 2\sup_{n \geq N}\,\abs{r_n} .$
Deduce that $\sum a_nx^n$ converges uniformly on $\intcc{0}{1}$ , that its sum is continuous there, and recover the radial limit of Exercise 11.9: $f(x) \to \sum a_n$ as $x \to 1^-$ .
(Abel’s theorem for Cauchy products) Let $\sum a_n = A$ , $\sum b_n = B$ and suppose the Cauchy product $\sum c_n$ , $c_n = \sum_{k} a_kb_{n-k}$ , converges, with sum $C$ . Prove $C = AB$ (inside the disk the product identity holds by Proposition 11.4; let $x \to 1^-$ ).
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 of two convergent series may diverge.
(A dividend of Exercise 11.5) 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 $x \to 1^-$ .

Show that $\sum (-1)^n$ is Abel-summable to $\frac12$ yet divergent: Abel’s theorem has no unconditional converse.
(Cesàro lemma) If $u_n \to 0$ then $\frac{u_1 + \dots + u_N}{N} \to 0$ (split the sum at a fixed $m$ ).
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 $\sum x_N^n \leq N$ ).
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 $L$ .
(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 $\sigma_N = \frac{A_0 + \dots + A_{N-1}}{N} \to L$ .

Show that a convergent series is Cesàro-summable to its sum (question 7 applied to $A_n - L$ ).
Compute the Cesàro value of $\sum(-1)^n$ and check it agrees with the Abel value $\frac12$ of question 6.
With $S_n = A_0 + \dots + A_n = (n+1)\,\sigma_{n+1}$ , prove the two identities, for $0 \leq x < 1$ :

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

Suppose $\sum a_nx^n$ converges on $\intco{0}{1}$ and $\sum \frac{a_n}{n+1}$ converges. Prove that the improper integral $\int_0^1 f$ 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 at $1$ by Part I).
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 convergent $\sum \frac1{n^2}$ , that $\eta = \frac12\sum_{n\geq1}\frac{1}{n^2}$ . (The Fourier chapter’s weekend problem evaluates $\sum\frac1{n^2} = \frac{\pi^2}{6}$ .)
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: $\frac{1}{1+x^3} = \frac{1/3}{1+x} + \frac{(2-x)/3}{x^2-x+1}$ ).
From the binomial series for $(1-t)^{-1/2}$ (Exercise 11.12) 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) — here not even Abel is needed.
(Catalan at the boundary) Show that $\sum C_n 4^{-n} = 2$ : the Catalan series of Example 11.21 converges at its radius $\frac14$ (asymptotics of Exercise 11.8), its sum is continuous on $\intcc{0}{\frac14}$ , and the closed form has limit $2$ there.

Part V — Rigidity: the identity theorem.

(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 $0$ .
(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 $0$ .
Find all functions $f$ analytic near $0$ with

$f\Bigl(\frac1k\Bigr) = \frac{k^2}{k^2+1} \qquad\text{for all large integers } k .$
Show that a function analytic on an open interval $I$ which vanishes on a subinterval vanishes identically on $I$ (the set of points around which $f$ vanishes identically is open and, by the identity theorem applied at accumulation points, closed in $I$ ). Conclude that no nonzero analytic function on $\R$ has compact support — whereas $C^\infty$ bump functions exist (Exercise 11.7 supplies the building block): analyticity is rigid, smoothness is flabby.
Synthesis. In one sentence each: (i) what Abel’s theorem adds to the normal-convergence package of Lemma 11.1; (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

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 there and its sum is continuous (Theorem 10.11). The value at $1$ being $\sum a_n$ , continuity at $1$ is the radial limit of Exercise 11.9.

3. For $\abs x < 1$ all three power series converge absolutely and $\bigl(\sum a_nx^n\bigr)\bigl(\sum b_nx^n\bigr) = \sum c_nx^n$ (Proposition 11.4). By question 2 each factor and the product side are continuous 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 diverges, although both factors converge (alternating series).

5. Exercise 11.5 gives $\frac{-\ln(1-x)}{1-x} = \sum H_nx^n$ ( $\abs x < 1$ ). Term-by-term primitives (Theorem 11.7 (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), 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); since $\sum\frac{a_n}{n+1}$ converges, Part I (question 2) makes $F$ continuous 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 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 at $0$ ). The series of $\frac{a_m}{m+1}$ is $\sum_{n\geq1}\frac{(-1)^{n-1}}{n^2}$ , absolutely convergent: question 16 gives $\int_0^1\frac{\ln(1+x)}{x}\dd x = \eta$ . In the absolutely 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 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), summable: the series converges normally on $\intcc{-1}{1}$ , its sum is continuous there, and coincides with the continuous $\arcsin$ 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): normal convergence of $\sum C_nx^n$ on $\intcc{0}{\frac14}$ , so its sum is continuous there; on $\intoo{0}{\frac14}$ it equals $\frac{1-\sqrt{1-4x}}{2x}$ (Example 11.21), 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 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 $f$ 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 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, so $Z = I$ : $f \equiv 0$ on $I$ . In particular an analytic function on $\R$ vanishing outside a compact set vanishes on an interval, hence everywhere: no nonzero analytic bump. The $C^\infty$ world is different: gluing the flat function of Exercise 11.7 (e.g. $x \mapsto \eu^{-1/x^2}\mathbf 1_{x>0}$ and its mirror) produces smooth bumps with compact support.

25. (i) Normal convergence lives on compact subdisks strictly inside the disk; Abel’s theorem extends 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 sits strictly between convergence and Abel summability (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 — same medicine, new patient.