---
title: "Sequences and Series"
book: "University Mathematics — Year 2"
subject: math
language: en
chapter: 7
exercises: 12
source: https://one-course.com/books/math/4/en/chapter/7-sequences-and-series
---

# Chapter 7 — Sequences and Series

The theory of numerical series (Year 1 volume) matures here in three directions: series with values in [Banach spaces](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-banach), where absolute convergence does the work; the finer tests for real series ([Abel summation](#thm-b2-series-abel)); and *[summable](#def-b2-series-summable) families* — summation freed from the order of the terms — with the Fubini theorem for double sums and the [Cauchy product](#thm-b2-series-fubini). These tools carry all the function-series chapters ahead.

## 7.1 Series in normed spaces

**Definition 7.1.**

For a sequence $(u_n)$ in a normed space $E$, the series $\sum u_n$ converges when its partial sums do; it converges *absolutely* when $\sum \norm{u_n} < \infty$. In a *Banach* space, absolute convergence implies convergence ([Theorem 5.21](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#thm-b2-nvs-absoluteconvergence)); in a [non-complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) space this may fail ([Exercise 7.9](#exo-b2-series-9)).

**Example 7.2.**

In $\mathcal{M}_n(K)$ (or $\mathcal{L}_c(E)$, $E$ Banach): for $\vertiii A < 1$, the *Neumann series* $\sum A^k$ converges [absolutely](#def-b2-series-def) to $(I - A)^{-1}$ (proved in [Exercise 5.5](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#exo-b2-nvs-5)); $\sum \frac{A^k}{k!}$ converges [absolutely](#def-b2-series-def) to $\eu^A$ for every $A$ ([Example 5.22](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ex-b2-nvs-matrixexp)). Operator-valued geometric and exponential series behave like their scalar models — the whole point of the Banach framework.

## 7.2 Abel summation

**Theorem 7.3 (Abel’s summation and test).**

(Summation by parts) For scalars $a_n$ and vectors $b_n$, with $B_n
= \sum_{k=0}^{n} b_k$:

$$
\sum_{n=0}^{N} a_n b_n
= a_N B_N - \sum_{n=0}^{N-1} (a_{n+1} - a_n) B_n .
$$

(Abel’s test) If $(a_n)$ is a real sequence, decreasing to $0$, and the partial sums $B_n$ are *bounded* (in a [Banach space](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-banach)), then $\sum a_n b_n$ converges.

**Proof.** The identity, step by step: with $B_{-1} = 0$, write $b_n = B_n -
B_{n-1}$ and split,

$$
\sum_{n=0}^{N} a_nb_n
= \sum_{n=0}^{N} a_nB_n - \sum_{n=0}^{N} a_nB_{n-1}
= \sum_{n=0}^{N} a_nB_n - \sum_{n=0}^{N-1} a_{n+1}B_{n} ,
$$

reindexing the second sum by $n \mapsto n + 1$ (the $B_{-1}$ term vanishes); collecting the common range $0 \leq n \leq N-1$ leaves $a_NB_N$ plus $\sum_{n\leq N-1}(a_n - a_{n+1})B_n$: the stated formula. It is the discrete integration by parts, with $(B_n)$ as antiderivative of $(b_n)$ and the difference $a_{n+1}
- a_n$ as derivative of $(a_n)$. For the test, with $\norm{B_n}
\leq M$: the boundary term $a_N B_N \to 0$; the series $\sum (a_n -
a_{n+1})B_n$ converges [absolutely](#def-b2-series-def), since

$$
\sum_n \norm{(a_{n+1} - a_n)B_n} \leq M \sum_n (a_n - a_{n+1})
= M a_0 < \infty
$$

(telescoping, $a_n \downarrow 0$). Both pieces of the identity converge, hence so does $\sum a_n b_n$. ∎

**Example 7.4.**

$\sum \frac{\sin n}{n}$ converges: $a_n = \frac1n \downarrow 0$ and $B_n = \sum_{k=1}^{n} \sin k$ is bounded — indeed $B_n =
\Im\sum_{k \leq n} \eu^{\iu k} = \Im\,\frac{\eu^{\iu}(\eu^{\iu n} -
1)}{\eu^{\iu} - 1}$, of modulus $\leq \frac{2}{\abs{\eu^{\iu} - 1}}$. It does *not* converge [absolutely](#def-b2-series-def) ($\abs{\sin n} \geq \sin^2 n
= \frac{1 - \cos 2n}{2}$, and $\sum \frac{1 - \cos 2n}{2n}$ diverges since $\sum \frac{\cos 2n}{n}$ converges by the same Abel test while $\sum \frac{1}{2n}$ diverges). The alternating series test is the special case $b_n = (-1)^n$.

**Example 7.5 (Abel on the circle of convergence).**

For which complex $z$ with $\abs z = 1$ does $\sum_{n \geq 1}
\frac{z^n}{n}$ converge? At $z = 1$ it is the harmonic series: divergent. For $z \neq 1$ on the circle, Abel’s test applies with $a_n = \frac1n \downarrow 0$ and $b_n = z^n$, whose partial sums are bounded independently of $N$:

$$
\Bigl|\sum_{n=1}^{N} z^n\Bigr|
= \Bigl|\frac{z(z^N - 1)}{z - 1}\Bigr|
\leq \frac{2}{\abs{z - 1}} .
$$

Convergent — though never [absolutely](#def-b2-series-def) ($\sum\frac1n$). One series, a circle of behaviours: divergence at a single point, semi-convergence everywhere else. This is the standard boundary behaviour of power series ([Chapter 11](https://one-course.com/books/math/4/en/chapter/11-power-series#ch-b2-powerseries)), met here with bare hands; at $z = -1$ it recovers the alternating harmonic series, and at $z = \eu^{\iu\theta}$ its real and imaginary parts are the series $\sum\frac{\cos
n\theta}{n}$ and $\sum\frac{\sin n\theta}{n}$ of [Exercise 7.4](#exo-b2-series-4).

**Example 7.6 (A booby-trapped alternating series).**

Does $\sum_{n\geq2} \dfrac{(-1)^n}{\sqrt n + (-1)^n}$ converge? The signs alternate and the terms tend to $0$ — yet the alternating test does *not* apply: the moduli $\frac{1}{\sqrt n + (-1)^n}$ are not decreasing (they jump up at each odd $n$). Expand instead:

$$
\frac{(-1)^n}{\sqrt n + (-1)^n}
= \frac{(-1)^n}{\sqrt n}\cdot
\frac{1}{1 + \frac{(-1)^n}{\sqrt n}}
= \frac{(-1)^n}{\sqrt n} - \frac{1}{n}
+ O\Bigl(\frac{1}{n^{3/2}}\Bigr).
$$

The first piece converges (alternating test, honestly applied to $\frac1{\sqrt n}\downarrow0$), the third converges [absolutely](#def-b2-series-def) — but the middle piece is the divergent harmonic series: the sum *diverges* to $-\infty$. The closing insight: when monotonicity fails, expand until every piece is either [absolutely](#def-b2-series-def) convergent or a clean test case; the hidden $-\frac1n$ is invisible to sign-counting.

**Remark 7.7 (Common pitfalls).**

(i) “Terms tend to $0$” proves nothing: the harmonic series diverges. (ii) The alternating test requires *decreasing* moduli — [Example 7.6](#ex-b2-series-trap) is the canonical counterexample, and [Exercise 7.1](#exo-b2-series-1)’s third series the drill. (iii) Conditionally convergent series may not be rearranged ([Example 7.12](#ex-b2-series-rearrange)), and their [Cauchy products](#thm-b2-series-fubini) may diverge: for $\sum\frac{(-1)^n}{\sqrt{n+1}}$ squared, the diagonal terms satisfy

$$
\abs{c_k} = \sum_{m=0}^{k}
\frac{1}{\sqrt{(m+1)(k-m+1)}}
\geq (k+1)\cdot\frac{2}{k+2} \longrightarrow 2 \neq 0
$$

(each factor is at most $\frac{k+2}2$ by AM–GM), so $\sum c_k$ diverges — absolute convergence of at least one factor ([Exercise 7.8](#exo-b2-series-8)) is not a luxury. (iv) [Summability](#def-b2-series-summable) is about *absolute* bounds by definition: there is no such thing as a conditionally [summable family](#def-b2-series-summable).

## 7.3 Summable families

**Definition 7.8.**

Let $I$ be a [countable](https://one-course.com/books/math/4/en/chapter/1-sets-and-structures#def-b2-structures-countable) index set. A family $(u_i)_{i \in I}$ of *nonnegative reals* is *summable* when the finite partial sums are bounded; its sum is

$$
\sum_{i \in I} u_i = \sup_{F \subseteq I \text{ finite}} \sum_{i
\in F} u_i \in \intcc{0}{+\infty} .
$$

A family of reals or complexes (or Banach vectors) is summable when $(\norm{u_i})$ is; its sum is then defined by splitting into positive/negative (or real/imaginary) parts — equivalently, as the common value of $\sum_{n} u_{\sigma(n)}$ over all enumerations $\sigma$ of $I$ (see below).

**Method 7.9 (Choosing a test).**

Facing $\sum u_n$, in order: (1) if $u_n \not\to 0$, divergence, stop. (2) If the terms have constant sign, compare: find an equivalent ([Chapter 6](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#ch-b2-comparison)) and place it on the Riemann–Bertrand map. (3) If signs alternate with *decreasing* moduli, the alternating test; if the moduli are not monotone, expand the term until each piece is [absolutely](#def-b2-series-def) convergent or a clean test case ([Example 7.6](#ex-b2-series-trap)). (4) If the sign pattern is oscillatory but structured ($\sin
n\theta$, $\eu^{\iu n\theta}$, matrix powers), Abel’s test with bounded partial sums. (5) Absolute convergence is always worth checking first: it is stronger, order-proof, and unlocks [Cauchy products](#thm-b2-series-fubini) and Fubini.

**Example 7.10 (Summability by diagonal counting).**

For which $s > 0$ is the family $\bigl((m + n)^{-s}\bigr)_{m, n \geq 1}$ [summable](#def-b2-series-summable)? Group the finite partial sums by diagonals $m + n = k$: the diagonal $k$ carries $k - 1$ pairs, each contributing $k^{-s}$, so the finite sums are exactly bounded by (and exhaust)

$$
\sum_{k \geq 2} \frac{k - 1}{k^{s}} ,
$$

a series with positive terms equivalent to $k^{1-s}$: [summable](#def-b2-series-summable) iff $s - 1 > 1$, i.e. $s > 2$. The two-dimensional index eats one full power: a plane of terms is “one dimension more divergent” than a line — the counting geometry of the index set, not the size of individual terms, decides [summability](#def-b2-series-summable). (The same census shows $\bigl((m^2 + n^2)^{-1}\bigr)$ is not [summable](#def-b2-series-summable): on the diagonal $m + n = k$, each term is at least $k^{-2}$, and $(k-1)\cdot k^{-2}$ sums like the harmonic series.)

**Theorem 7.11 (Summability and order).**

1. For nonnegative families, the sum is invariant under any enumeration: $\sum_{i} u_i = \sum_{n=0}^{\infty}  u_{\sigma(n)}$ for every bijection $\sigma \colon \N \to I$ .
2. A real or complex series $\sum u_n$ is *commutatively convergent* (every rearrangement converges, with the same sum) if and only if it is [absolutely](#def-b2-series-def) convergent.

**Proof.** (1) Every partial sum $\sum_{n \leq N} u_{\sigma(n)}$ is a finite partial sum of the family (so $\leq$ the sup); every finite $F$ is contained in some $\{\sigma(0), \dots, \sigma(N)\}$ (so the sup $\leq$ the series’ limit). The two bounds match.

(2) If $\sum\abs{u_n} < \infty$: for any rearrangement $\sigma$ and $\varepsilon > 0$, choose $N$ with $\sum_{n > N}\abs{u_n} \leq
\varepsilon$; beyond the rank where $\sigma$ has exhausted $\intint{0}{N}$, the rearranged partial sums differ from the original limit by at most $\varepsilon$: same sum. If $\sum
\abs{u_n} = \infty$ but $\sum u_n$ converges (real case; complex follows coordinatewise): the positive and negative parts both diverge, and one can rearrange to reach any prescribed limit — Riemann’s theorem, carried out in [Exercise 7.5](#exo-b2-series-5) — so commutative convergence fails. ∎

**Example 7.12 (A rearrangement caught red-handed).**

The alternating harmonic series sums to $\sum_{n\geq1}\frac{(-1)^{n-1}}{n} = \ln 2$ (Year 1 volume). Rearrange it as “one positive, two negatives”:

$$
1 - \frac12 - \frac14 + \frac13 - \frac16 - \frac18 + \frac15 -
\cdots
$$

Grouping each block of three,

$$
\frac{1}{2k-1} - \frac{1}{4k-2} - \frac1{4k}
= \frac{1}{4k-2} - \frac{1}{4k}
= \frac12\Bigl(\frac{1}{2k-1} - \frac1{2k}\Bigr),
$$

so the rearranged series converges to $\frac12\ln 2$ — half the original sum, with exactly the same terms. [Non-absolutely](#def-b2-series-def) convergent series remember the order of their terms; [summable](#def-b2-series-summable) families are precisely the ones that do not.

**Example 7.13 (Grouping is safe, ungrouping is not).**

Grouping consecutive terms of a *convergent* series never changes the sum: the grouped partial sums form a subsequence of the original ones. The converse operation is forbidden:

$$
(1 - 1) + (1 - 1) + (1 - 1) + \cdots = 0 + 0 + \cdots = 0,
$$

yet the ungrouped $1 - 1 + 1 - 1 + \cdots$ diverges (partial sums oscillate between $1$ and $0$). Ungrouping is legitimate only with a compensating hypothesis — for instance, terms tending to $0$ with bounded group lengths: then between two grouped partial sums the original ones drift by at most a sum of $o(1)$ terms of bounded number, and convergence transfers back. That is exactly the clause under which the block computation of [Example 7.12](#ex-b2-series-rearrange) is a proof and not a sleight of hand.

**Theorem 7.14 (Fubini for families; Cauchy products).**

Let $(u_{m,n})_{(m,n) \in \N^2}$ be a [summable](#def-b2-series-summable) double family (i.e. $\sup_F \sum_F \abs{u_{m,n}} < \infty$). Then

$$
\sum_{(m,n)} u_{m,n}
= \sum_{m=0}^{\infty}\Bigl(\sum_{n=0}^{\infty} u_{m,n}\Bigr)
= \sum_{n=0}^{\infty}\Bigl(\sum_{m=0}^{\infty} u_{m,n}\Bigr)
= \sum_{k=0}^{\infty}\Bigl(\sum_{m+n=k} u_{m,n}\Bigr),
$$

all inner series converging ([absolutely](#def-b2-series-def)). In particular, if $\sum
a_m$ and $\sum b_n$ converge [absolutely](#def-b2-series-def), their *Cauchy product* converges [absolutely](#def-b2-series-def) with

$$
\Bigl(\sum_m a_m\Bigr)\Bigl(\sum_n b_n\Bigr)
= \sum_{k=0}^{\infty} c_k,
\qquad
c_k = \sum_{m=0}^{k} a_m b_{k-m} .
$$

**Proof.** *Nonnegative case.* Each grouping (by rows, columns, or diagonals) computes the same supremum: any finite set of pairs is contained in a finite block of rows (bounding each grouped sum below by finite partial sums and above by the total), and monotone convergence of partial sums does the rest — concretely, for rows: $\sum_{m \leq M}\sum_{n \leq N} u_{m,n} \leq S$ gives, letting $N
\to \infty$ then $M \to \infty$, $\sum_m \sum_n u_{m,n} \leq S$; conversely every finite $F$ sits in such a rectangle, so $S \leq
\sum_m\sum_n u_{m,n}$. Diagonals: same two bounds with triangles instead of rectangles.

*General case.* Split into positive and negative (real and imaginary) parts, each a [summable](#def-b2-series-summable) nonnegative family; the four groupings agree on each part, hence on the difference; absolute convergence of the inner series comes from the nonnegative case applied to $\abs{u_{m,n}}$.

*[Cauchy product](#thm-b2-series-fubini).* The family $u_{m,n} = a_m b_n$ is [summable](#def-b2-series-summable): finite partial sums of $\abs{a_mb_n}$ are bounded by $\bigl(\sum\abs{a_m}\bigr)\bigl(\sum\abs{b_n}\bigr)$. Rows give $\bigl(\sum a_m\bigr)\bigl(\sum b_n\bigr)$; diagonals give $\sum_k
c_k$. ∎

**Example 7.15 (The exponential identity, honestly).**

For $a, b \in \C$ (or commuting matrices):

$$
\Bigl(\sum_m \frac{a^m}{m!}\Bigr)\Bigl(\sum_n
\frac{b^n}{n!}\Bigr)
= \sum_k \sum_{m+n=k} \frac{a^m b^n}{m!\,n!}
= \sum_k \frac{(a + b)^k}{k!},
$$

by the binomial theorem on each diagonal: $\eu^a \eu^b = \eu^{a+b}$ — the functional equation of $\exp$ derived from the series alone. (Commutation is used in the binomial step; for non-commuting matrices the identity genuinely fails, [Chapter 16](https://one-course.com/books/math/4/en/chapter/16-differential-equations#ch-b2-diffeq).)

**Example 7.16 (Cauchy products as a computing device).**

From the geometric series and [Exercise 7.2](#exo-b2-series-2)’s $\sum_{n
\geq 1} nz^n = \frac{z}{(1-z)^2}$ ($\abs z < 1$), one more [Cauchy product](#thm-b2-series-fubini) finishes the second moment. Multiply $\sum_m
mz^m$ by $\sum_n z^n$: the diagonal coefficient is $\sum_{m=0}^k
m = \frac{k(k+1)}2$, so

$$
\frac{z}{(1-z)^3} = \sum_{k\geq0}\frac{k(k+1)}{2}\,z^k ,
$$

and the identity $n^2 = 2\cdot\frac{n(n+1)}2 - n$ assembles

$$
\sum_{n\geq1} n^2z^n = \frac{2z}{(1-z)^3} - \frac{z}{(1-z)^2}
= \frac{z(1+z)}{(1-z)^3} .
$$

At $z = \frac12$: $\sum_{n\geq1}\frac{n^2}{2^n} =
\frac{\frac12\cdot\frac32}{\frac18} = 6$ — a closed value with no differentiation anywhere, just [absolutely](#def-b2-series-def) convergent series multiplied like polynomials. The same telescoping of identities computes every $\sum n^dz^n$, and probabilists will recognize the second factorial moment of the geometric law ([Chapter 23](https://one-course.com/books/math/4/en/chapter/23-probability-generating-functions#ch-b2-genfun)).

**Example 7.17 (A double-sum evaluation).**

For $s > 1$ real, let $\zeta(s) = \sum_{n\geq1} n^{-s}$. Counting divisors by double summation — the family $(m^{-s}n^{-s})$ over $(m,n) \in (\N^*)^2$ is [summable](#def-b2-series-summable) (product of convergent positive series) — and grouping by the product $q = mn$:

$$
\zeta(s)^2 = \sum_{m,n} \frac{1}{(mn)^s}
= \sum_{q=1}^{\infty} \frac{d(q)}{q^s},
$$

where $d(q)$ is the number of divisors of $q$. [Summable](#def-b2-series-summable) families turn combinatorics into analysis.

**Example 7.18 (A Fubini evaluation: ∑n(ζ(n)−1)=1\sum_n (\zeta(n) - 1) = 1∑n​(ζ(n)−1)=1).**

For integer $n \geq 2$, $\zeta(n) - 1 = \sum_{k \geq 2} k^{-n}$. The double family $(k^{-n})_{k, n \geq 2}$ is [summable](#def-b2-series-summable): summing the geometric columns first,

$$
\sum_{k\geq2}\sum_{n\geq2} \frac{1}{k^n}
= \sum_{k\geq2} \frac{1/k^2}{1 - 1/k}
= \sum_{k\geq2} \frac{1}{k(k-1)} = 1
$$

(telescoping), and all terms are positive, so [Theorem 7.14](#thm-b2-series-fubini) authorizes summing by rows instead:

$$
\sum_{n\geq2}\bigl(\zeta(n) - 1\bigr) = 1 .
$$

The infinitely many $\zeta$-values, each transcendental-looking, have tails that add up to exactly $1$. Closing insight: when a double sum has positive terms, compute it in whichever order collapses — here columns are geometric, rows are mysterious, and Fubini transfers the collapse.

**Example 7.19 (The geometric series solves an equation).**

In the [Banach space](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-banach) $\bigl(C(\intcc01),
\norm\cdot_\infty\bigr)$, solve $x - K(x) = y$ where $K(f)$ is the constant function $\frac12\int_0^1 f$. The [operator norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#thm-b2-nvs-continuouslinear) is $\vertiii K \leq \frac12 < 1$, so the Neumann series applies ([Example 7.2](#ex-b2-series-neumann)): $x = \sum_{n\geq0} K^n(y)$. Compute the iterates: $K(y) = \frac12\int_0^1 y$ (a constant), and applying $K$ to a constant $c$ gives $\frac c2$, so $K^n(y) = \frac{1}{2^{n-1}}\cdot\frac12\int_0^1 y$ for $n \geq
1$. Summing the geometric constants:

$$
x = y + \Bigl(\int_0^1 y\Bigr)
\sum_{n\geq1}\frac{1}{2^n} = y + \int_0^1 y .
$$

Check: $x - K(x) = y + \int y - \frac12\bigl(\int y + \int
y\bigr) = y$. An infinite series, a finite answer, and a one-line verification — the geometric series is an inversion algorithm, not just a convergence statement.

**Example 7.20 (Telescoping by partial fractions).**

Exact summation is rare; telescoping is its main supplier. Decompose

$$
\frac{1}{n(n+1)(n+2)}
= \frac{1}{2}\Bigl(\frac{1}{n(n+1)} -
\frac{1}{(n+1)(n+2)}\Bigr),
$$

(check by reduction to the common denominator), so the partial sums collapse:

$$
\sum_{n=1}^{N}\frac{1}{n(n+1)(n+2)}
= \frac12\Bigl(\frac{1}{1\cdot2} -
\frac{1}{(N+1)(N+2)}\Bigr)
\longrightarrow \frac14 .
$$

The same pattern — write the term as $c(u_n - u_{n+1})$ for an explicit $(u_n)$ — solved [Exercise 7.10](#exo-b2-series-10) (arctangents) and computes every $\sum\frac{1}{n(n+1)\cdots(n +
k)} = \frac{1}{k\cdot k!}$. When an exact sum exists at this level, a telescope is usually hiding in the term.

**Remark 7.21 (Perspectives within this volume).**

Three chapters ahead are direct clients. For [Chapter 10](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#ch-b2-funcseq): normal convergence of $\sum f_n$ is absolute convergence of $\sum\norm{f_n}_\infty$ in the [Banach space](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-banach) $\bigl(C, \norm\cdot_\infty\bigr)$ — this chapter’s [Theorem 5.21](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#thm-b2-nvs-absoluteconvergence) in costume. For [Chapter 11](https://one-course.com/books/math/4/en/chapter/11-power-series#ch-b2-powerseries): inside the disk of convergence everything is absolute and [summable](#def-b2-series-summable), so [Cauchy products](#thm-b2-series-fubini) and rearrangements run free (that is why power series multiply like polynomials); on the boundary, Abel’s test takes over ([Example 7.5](#ex-b2-series-abelboundary)). For [Chapter 23](https://one-course.com/books/math/4/en/chapter/23-probability-generating-functions#ch-b2-genfun): probability generating functions are power series whose manipulations — products for sums of independent variables, double sums for compound laws — are all licensed by [Theorem 7.14](#thm-b2-series-fubini). [Summable](#def-b2-series-summable) families are the legal department of the analysis to come.

**Remark 7.22 (Where this chapter is used).**

Everything with an infinite sum passes through here: power series ([Chapter 11](https://one-course.com/books/math/4/en/chapter/11-power-series#ch-b2-powerseries)) are [summable](#def-b2-series-summable) families in disguise, Fourier coefficients get multiplied by [Cauchy products](#thm-b2-series-fubini) and rearranged by Parseval ([Chapter 14](https://one-course.com/books/math/4/en/chapter/14-fourier-series#ch-b2-fourier)), and probability generating functions ([Chapter 23](https://one-course.com/books/math/4/en/chapter/23-probability-generating-functions#ch-b2-genfun)) are Fubini’s theorem applied to expectations. The Year 3 volume absorbs [summable](#def-b2-series-summable) families into Lebesgue integration over the counting measure — where [Theorem 7.14](#thm-b2-series-fubini) becomes a special case of the Fubini–Tonelli theorem.

## 7.4 Exercises

**Exercise 7.1 ★.**

Nature of: $\sum \dfrac{\cos n}{n}$; $\;\sum \dfrac{(-1)^n}{\ln
n}$; $\;\sum \dfrac{(-1)^n}{n^{3/4} + \cos n}$ *(expand as in the Year 1 trap: the alternating test needs monotonicity)*.

**Solution of Exercise 7.1.**

$\sum\frac{\cos n}{n}$: Abel’s test with $a_n = \frac1n$ and $b_n =
\cos n$, whose partial sums are bounded (real part of a geometric sum, as in [Example 7.4](#ex-b2-series-sinn)): convergent (not [absolutely](#def-b2-series-def), by the same $\cos^2$ trick).

$\sum \frac{(-1)^n}{\ln n}$ ($n \geq 2$): alternating test, $\frac{1}{\ln n} \downarrow 0$: convergent; not [absolutely](#def-b2-series-def) ($\ln n
\leq n$).

$\sum \frac{(-1)^n}{n^{3/4} + \cos n}$: expand,

$$
\frac{(-1)^n}{n^{3/4} + \cos n}
= \frac{(-1)^n}{n^{3/4}}\cdot
\frac{1}{1 + \frac{\cos n}{n^{3/4}}}
= \frac{(-1)^n}{n^{3/4}}
- \frac{(-1)^n\cos n}{n^{3/2}} + O\Bigl(\frac{1}{n^{9/4}}\Bigr).
$$

First series: alternating, convergent. Second: [absolutely](#def-b2-series-def) convergent ($\frac{1}{n^{3/2}}$ scale). Third: [absolutely](#def-b2-series-def) convergent. Total: convergent.

**Exercise 7.2 ★.**

Prove that for $\abs z < 1$: $\sum_{n\geq1} n z^{n} =
\dfrac{z}{(1-z)^2}$, via the [Cauchy product](#thm-b2-series-fubini) of $\sum z^n$ with itself.

**Solution of Exercise 7.2.**

[Cauchy product](#thm-b2-series-fubini) of $\sum_{m\geq0} z^m$ with itself (both [absolutely](#def-b2-series-def) convergent for $\abs z < 1$): the diagonal coefficient is $c_k =
\sum_{m=0}^{k} 1 = k + 1$, so

$$
\frac{1}{(1-z)^2} = \sum_{k\geq0} (k+1)z^k .
$$

Multiplying by $z$ and reindexing: $\sum_{n \geq 1} n z^n =
\frac{z}{(1-z)^2}$.

**Exercise 7.3 ★★.**

(Kronecker-type lemma) Let $\sum b_n$ be a convergent real series. Prove, by [Abel summation](#thm-b2-series-abel), that $\dfrac{1}{n}\sum_{k=1}^{n} k\,b_k
\to 0$.

**Solution of Exercise 7.3.**

Let $B_n = \sum_{k \leq n} b_k \to B$. [Abel summation](#thm-b2-series-abel) with $a_k = k$:

$$
\sum_{k=1}^{n} k\,b_k = n B_n - \sum_{k=1}^{n-1} B_k
\quad\Longrightarrow\quad
\frac1n \sum_{k=1}^{n} k b_k = B_n - \frac{1}{n}\sum_{k=1}^{n-1}
B_k .
$$

The Cesàro means of the convergent $(B_k)$ tend to its limit $B$ (Year 1 volume), so the right side tends to $B - B = 0$.

**Exercise 7.4 ★★.**

Study the convergence of $\sum \dfrac{\sin(n\theta)}{n^\alpha}$ ($\theta \in \R$, $\alpha > 0$) — for which $(\theta, \alpha)$ is it [absolutely](#def-b2-series-def) convergent, semi-convergent, divergent?

**Solution of Exercise 7.4.**

If $\theta \in \pi\Z$: all terms vanish — convergent trivially. Assume $\theta \notin \pi\Z$.

$\alpha > 1$: [absolutely](#def-b2-series-def) convergent (domination by $n^{-\alpha}$).

$0 < \alpha \leq 1$: Abel’s test applies ($a_n = n^{-\alpha}
\downarrow 0$; partial sums of $\sin n\theta$ bounded by $\frac{1}{\abs{\sin(\theta/2)}}$, geometric sum): convergent. Not [absolutely](#def-b2-series-def): $\abs{\sin n\theta} \geq \sin^2 n\theta = \frac{1 -
\cos 2n\theta}{2}$, and $\sum \frac{1 - \cos 2n\theta}{2n^\alpha}$ diverges ($\sum n^{-\alpha}$ diverges; $\sum
\frac{\cos 2n\theta}{n^\alpha}$ converges by Abel when $2\theta
\notin 2\pi\Z$; the excluded case $2\theta \in 2\pi\Z$ means $\theta \in \pi\Z$, already handled). Semi-convergent.

**Exercise 7.5 ★★★.**

(Riemann rearrangement) Let $\sum u_n$ be a convergent but not [absolutely](#def-b2-series-def) convergent real series, and $\ell \in \R$. Prove that some rearrangement of $\sum u_n$ converges to $\ell$. *(Show both subseries of positive and negative terms diverge; then greedily alternate: take positive terms until exceeding $\ell$, then negative until dropping below, and so on; the terms tend to $0$, forcing convergence to $\ell$.)*

**Solution of Exercise 7.5.**

Let $p_1, p_2, \dots$ be the nonnegative terms of $(u_n)$ in order, $q_1, q_2, \dots$ the negative ones. Both $\sum p_k$ and $\sum
q_k$ diverge: if one of them converged, the other would equal the convergent $\sum u_n$ minus it, hence converge too — and then $\sum \abs{u_n} = \sum p_k - \sum q_k$ would converge, contradicting the hypothesis. Also $u_n \to 0$ ($\sum u_n$ converges).

Greedy rearrangement: take positive terms $p_1, p_2, \dots$ until the running total first exceeds $\ell$ (possible: $\sum p_k =
+\infty$); then negative terms until the total first drops below $\ell$ (possible: $\sum q_k = -\infty$); repeat forever (each phase is finite, and every term is used exactly once: a genuine rearrangement). After each switch, the distance from the running total to $\ell$ is at most the last term used; since the terms used at the $m$-th switch have index $\to \infty$, and $u_n \to 0$, the running totals converge to $\ell$.

**Exercise 7.6 ★★.**

Prove that the family $\Bigl(\dfrac{x^{m+n}}{m!\,n!}\Bigr)_{(m,n)\in\N^2}$ is [summable](#def-b2-series-summable) for every $x \in \R$, and re-derive the identity $(\eu^x)^2 =
\eu^{2x}$ by grouping the double sum along the diagonals $m + n =
k$.

**Solution of Exercise 7.6.**

[Summability](#def-b2-series-summable): the finite partial sums of $\frac{\abs x^{m+n}}{m!n!}$ are bounded by $\bigl(\sum_m
\frac{\abs x^m}{m!}\bigr)^2 = \eu^{2\abs x}$. Diagonal grouping ([Theorem 7.14](#thm-b2-series-fubini)):

$$
(\eu^{x})^2 = \sum_{m,n} \frac{x^{m+n}}{m!\,n!}
= \sum_{k=0}^{\infty} x^k \sum_{m+n=k} \frac{1}{m!\,n!}
= \sum_k \frac{x^k}{k!}\sum_{m=0}^{k}\binom km
= \sum_k \frac{(2x)^k}{k!} = \eu^{2x} .
$$

**Exercise 7.7 ★★.**

Prove that the family $\bigl(\frac{1}{m^2 n^2}\bigr)_{m,n \geq 1}$ is [summable](#def-b2-series-summable), and that grouping by $\gcd$: with $q = \gcd(m,n)$,

$$
\zeta(2)^2 = \sum_{q\geq1} \frac{1}{q^4}
\sum_{\substack{a,b \geq 1\\ \gcd(a,b)=1}} \frac{1}{a^2b^2}
= \zeta(4) \cdot S,
$$

where $S = \sum_{\gcd(a,b)=1} \frac{1}{a^2b^2}$: deduce $S =
\zeta(2)^2/\zeta(4)$. *(Every pair $(m,n)$ writes uniquely $(qa, qb)$ with $\gcd(a,b) = 1$.)*

**Solution of Exercise 7.7.**

[Summability](#def-b2-series-summable): bounded by $\zeta(2)^2$ as a product family ([Theorem 7.14](#thm-b2-series-fubini)’s Cauchy-product argument). The map $(q, a, b) \mapsto (qa, qb)$, from triples with $\gcd(a, b) = 1$ to pairs $(m, n)$, is a bijection (set $q = \gcd(m,n)$). Grouping the [summable family](#def-b2-series-summable) accordingly (a partition of the index set — legitimate for [summable](#def-b2-series-summable) families by [Theorem 7.11](#thm-b2-series-rearrangement)/[Theorem 7.14](#thm-b2-series-fubini) applied to the partition into [countably](https://one-course.com/books/math/4/en/chapter/1-sets-and-structures#def-b2-structures-countable) many classes):

$$
\zeta(2)^2 = \sum_{q} \sum_{\gcd(a,b)=1} \frac{1}{q^4 a^2 b^2}
= \zeta(4)\, S,
\qquad\text{so}\qquad
S = \frac{\zeta(2)^2}{\zeta(4)} .
$$

(With the values $\zeta(2) = \frac{\pi^2}{6}$, $\zeta(4) = \frac{\pi^4}{90}$ from [Chapter 14](https://one-course.com/books/math/4/en/chapter/14-fourier-series#ch-b2-fourier): $S = \frac{5}{2}$.)

**Exercise 7.8 ★★★.**

(Abel’s theorem on products, light version) Suppose $\sum a_n$ converges [absolutely](#def-b2-series-def) and $\sum b_n$ converges. Prove that their [Cauchy product](#thm-b2-series-fubini) $\sum c_n$ converges, with $\sum c_n = (\sum
a_n)(\sum b_n)$. *(Write $C_N = \sum_{k\leq N} c_k = \sum_n
a_n B_{N-n}$ with $B$ the partial sums of $b$; split according to $n \leq N/2$ or not, using boundedness of $(B_m)$ and the absolute tail of $(a_n)$.)*

**Solution of Exercise 7.8.**

Let $A = \sum a_n$ (absolute), $B_m = \sum_{k\leq m} b_k \to B$, bounded by $M$. Then

$$
C_N = \sum_{k=0}^{N} c_k = \sum_{n=0}^{N} a_n B_{N-n}
$$

(collect by the index of $a$). Write

$$
C_N - AB = \sum_{n=0}^{N} a_n (B_{N-n} - B) - B\sum_{n > N} a_n .
$$

The last term tends to $0$. Split the sum at $n = \lfloor N/2
\rfloor$: for $n \leq N/2$, $N - n \geq N/2$, so $\abs{B_{N-n} - B}
\leq \varepsilon_N := \sup_{m \geq N/2}\abs{B_m - B} \to 0$, and this part is $\leq \varepsilon_N \sum\abs{a_n}$; for $n > N/2$, $\abs{B_{N-n} - B} \leq 2M$, and this part is $\leq 2M \sum_{n >
N/2} \abs{a_n} \to 0$. Hence $C_N \to AB$.

**Exercise 7.9 ★★★.**

In the ([non-complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete)) space $E$ of eventually-zero real sequences with the sup [norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm), exhibit an [absolutely](#def-b2-series-def) convergent series that does not converge in $E$. *(Try $u_n = 2^{-n} e_n$ with $(e_n)$ the canonical sequences.)*

**Solution of Exercise 7.9.**

Take $u_n = 2^{-n} e_n$ ($e_n$ the sequence with a single $1$ in position $n$). Then $\sum \norm{u_n}_\infty = \sum 2^{-n} <
\infty$: [absolutely](#def-b2-series-def) convergent. But the partial sums $S_N =
(1, \tfrac12, \dots, 2^{-N}, 0, \dots)$ would have to converge to the sequence $(2^{-n})_n$, which is *not* eventually zero: outside $E$. Inside $E$, $(S_N)$ is Cauchy without limit ($\norm{S_N
- x}_\infty \geq 2^{-N-1}$ fails to help any eventually-zero $x$: for any $x \in E$ vanishing beyond rank $K$, $\norm{S_N - x} \geq
2^{-K-1}$ for $N > K$): the series does not converge in $E$. [Completeness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) is exactly what [Theorem 5.21](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#thm-b2-nvs-absoluteconvergence) needs.

**Exercise 7.10 ★★.**

Verify the identity $\arctan(n+1) - \arctan(n) =
\arctan\dfrac{1}{n^2 + n + 1}$, and deduce the exact value of

$$
\sum_{n=1}^{\infty} \arctan\frac{1}{n^2 + n + 1} .
$$

**Solution of Exercise 7.10.**

Both $\arctan(n+1) - \arctan n$ and $\arctan\frac{1}{n^2+n+1}$ lie in $\intoo{0}{\frac\pi2}$, and the tangent addition formula gives

$$
\tan\bigl(\arctan(n{+}1) - \arctan n\bigr)
= \frac{(n+1) - n}{1 + n(n+1)} = \frac{1}{n^2 + n + 1} :
$$

equal tangents in an interval where $\tan$ is injective, so the identity holds. Telescoping,

$$
\sum_{n=1}^{N}\arctan\frac{1}{n^2+n+1}
= \arctan(N{+}1) - \arctan 1 \xrightarrow[N\to\infty]{}
\frac\pi2 - \frac\pi4 = \frac\pi4 .
$$

**Exercise 7.11 ★★.**

Determine the nature (with equivalents) of

$$
\sum_n \bigl(\sqrt{n+1} - \sqrt n\bigr)^{\alpha}
\ (\alpha > 0), \qquad
\sum_n \Bigl(1 - \cos\frac1n\Bigr), \qquad
\sum_n \Bigl(\eu - \Bigl(1 + \frac1n\Bigr)^{\!n}\Bigr).
$$

**Solution of Exercise 7.11.**

First: $\sqrt{n+1} - \sqrt n = \frac{1}{\sqrt{n+1} + \sqrt n}
\sim \frac{1}{2\sqrt n}$, so the terms are $\sim
2^{-\alpha}n^{-\alpha/2}$: convergence iff $\frac\alpha2 > 1$, i.e. $\alpha > 2$. Second: $1 - \cos\frac1n \sim
\frac{1}{2n^2}$: converges. Third: $\bigl(1 + \frac1n\bigr)^n =
\eu^{\,n\ln(1 + 1/n)} = \eu^{\,1 - \frac1{2n} + O(n^{-2})} =
\eu\bigl(1 - \frac{1}{2n} + O(n^{-2})\bigr)$, so

$$
\eu - \Bigl(1 + \frac1n\Bigr)^{\!n} \sim \frac{\eu}{2n} :
$$

positive terms equivalent to a harmonic multiple: diverges.

**Exercise 7.12 ★★★.**

Let $(a_n)$ be positive and *decreasing* with $\sum a_n$ convergent. Prove that $n\,a_n \to 0$ *(bound $n a_{2n}$ by a tail)*. Show that the converse fails, and that monotonicity is essential, with explicit counterexamples.

**Solution of Exercise 7.12.**

By monotonicity, $n\,a_{2n} \leq a_{n+1} + a_{n+2} + \dots +
a_{2n} = S_{2n} - S_n \to 0$ (Cauchy criterion for the convergent series). Hence $2n\,a_{2n} \to 0$, and $(2n{+}1)\,
a_{2n+1} \leq (2n{+}1)a_{2n} = \frac{2n+1}{2n}\,(2n\,a_{2n}) \to
0$: both subsequences of $(na_n)$ tend to $0$, so $na_n \to 0$.

*Converse fails:* $a_n = \frac1{n\ln n}$ is positive decreasing with $na_n = \frac1{\ln n} \to 0$, yet $\sum a_n$ diverges (Bertrand frontier, [Problem 7.1](#pb-b2-series-1), question 18). *Monotonicity essential:* let $a_n = \frac1n$ when $n$ is a power of $2$ and $a_n = 2^{-n}$ otherwise: $\sum a_n \leq
\sum_k 2^{-k} + \sum_n 2^{-n} < \infty$, but $na_n = 1$ along the powers of $2$: $na_n \not\to 0$.

## 7.5 Problem: Euler’s $\zeta(2) = \pi^2/6$, by Cauchy’s Cotangent Sum

Euler’s most famous identity, $1 + \frac14 + \frac19 + \cdots =
\frac{\pi^2}6$, admits a completely elementary proof, due to Cauchy: de Moivre’s formula produces a polynomial whose roots are the numbers $\cot^2\frac{k\pi}{2n+1}$, Vieta sums those roots exactly, and the squeeze $\cot^2\theta < \frac{1}{\theta^2} <
1 + \cot^2\theta$ crushes the partial sums of $\sum\frac1{k^2}$ between two explicit rational bounds. We run the proof in full, extract $\zeta(4) = \frac{\pi^4}{90}$ by the same method, then map the entire frontier between convergence and divergence with the Bertrand series — and prove that the frontier carries no slowest convergent series at all.

**Problem 7.1.**

Weekend problem — $\zeta(2) = \pi^2/6$ and the Bertrand panorama

Throughout, $n \geq 1$ and $\theta_k = \dfrac{k\pi}{2n+1}$ for $k
= 1, \dots, n$; note $0 < \theta_k < \frac\pi2$.

**Part I — The cotangent identity.**

1. Prove de Moivre’s formula $(\cos\theta +  \iu\sin\theta)^m = \cos m\theta + \iu\sin m\theta$ ($m  \in \N$), and deduce, for $m = 2n + 1$, $$\sin\bigl((2n{+}1)\theta\bigr) = \sum_{j=0}^{n}  (-1)^j\binom{2n+1}{2j+1}\cos^{2(n-j)}\theta\,  \sin^{2j+1}\theta .$$
2. Deduce that for $\theta \in \intoo{0}{\frac\pi2}$, $$\sin\bigl((2n{+}1)\theta\bigr) =  \sin^{2n+1}\theta\; P_n(\cot^2\theta),  \qquad  P_n(x) = \sum_{j=0}^{n}  (-1)^j\binom{2n+1}{2j+1}x^{\,n-j},$$ a polynomial of degree $n$ with leading coefficient $2n  + 1$.
3. Show that $x_k = \cot^2\theta_k$ , $k = 1, \dots, n$ , are $n$ *distinct* roots of $P_n$ — hence all of them.
4. By Vieta, prove the exact identity $$\sum_{k=1}^{n} \cot^2\frac{k\pi}{2n+1}  = \frac{n(2n-1)}{3}.$$
5. Deduce also $\displaystyle\sum_{k=1}^{n}  \frac{1}{\sin^2\theta_k} = \frac{2n(n+1)}{3}$ .
6. Prove the squeeze: $\cot^2\theta < \dfrac1{\theta^2} <  \dfrac{1}{\sin^2\theta}$ for $\theta \in  \intoo{0}{\frac\pi2}$ *(from $\sin\theta < \theta <  \tan\theta$)* .

**Part II — The squeeze closes: Euler’s theorem.**

7. Summing question 6 over $k = 1, \dots, n$ with $\theta =  \theta_k$, establish $$\frac{n(2n-1)}{3} \;<\;  \frac{(2n+1)^2}{\pi^2}\sum_{k=1}^{n}\frac1{k^2}  \;<\; \frac{2n(n+1)}{3}.$$
8. Conclude (*Euler’s theorem, by Cauchy’s proof*): $$\zeta(2) = \sum_{k=1}^{\infty}\frac{1}{k^2} =  \frac{\pi^2}{6}.$$
9. Extract a rate from the sandwich: show $$\Bigl|\sum_{k=1}^{n}\frac1{k^2} - \frac{\pi^2}6\Bigr|  = O\Bigl(\frac1n\Bigr),$$ consistent with the exact tail $\sum_{k>n}k^{-2} =  \frac1n - \frac{1}{2n^2} + O(n^{-3})$ of [Exercise 6.11](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#exo-b2-comparison-11).
10. Run the machine one floor higher: using the second Vieta function of $P_n$, show $$\sum_{k=1}^{n}\cot^4\theta_k =  \Bigl(\frac{n(2n-1)}3\Bigr)^{\!2} -  \frac{2n(2n-1)(2n-2)(2n-3)}{60}  \;\sim\; \frac{8n^4}{45},$$ and squeeze with $\cot^4 < \theta^{-4} < (1 +  \cot^2)^2$ to obtain $\zeta(4) = \dfrac{\pi^4}{90}$.

**Part III — Dividends.**

11. Deduce from $\zeta(2) = \frac{\pi^2}6$: $$\sum_{k\geq0}\frac{1}{(2k+1)^2} = \frac{\pi^2}{8},  \qquad  \sum_{k\geq1}\frac{(-1)^{k-1}}{k^2} = \frac{\pi^2}{12}.$$
12. Combine with [Exercise 7.7](#exo-b2-series-7) : compute $S =  \sum_{\gcd(a,b)=1}\frac{1}{a^2b^2} =  \frac{\zeta(2)^2}{\zeta(4)} = \frac52$ , and interpret $\frac{1}{\zeta(2)} = \frac{6}{\pi^2} \approx 0.608$ as the density of coprime pairs (state the heuristic honestly: the rigorous count is a Year 3 volume matter).
13. (Certified acceleration) The tail formula of question 9 gives $\sum_{k\leq n}k^{-2} + \frac1n - \frac1{2n^2} =  \frac{\pi^2}6 + O(n^{-3})$ . Compare the work needed for six digits of $\zeta(2)$ : direct summation versus the corrected sum at $n = 100$ (where the error is $1.7\cdot10^{-7}$ ).
14. Check question 4 by hand at $n = 1$ and $n = 2$ (the values $\cot^2\frac\pi3 = \frac13$ and $\cot^2\frac\pi5 + \cot^2\frac{2\pi}5 = 2$ ), using $\cos\frac\pi5 = \frac{1+\sqrt5}4$ or a numerical evaluation.
15. Prove the companion identity $$\sum_{k=1}^{n}\tan^2\frac{k\pi}{2n+1} = n(2n+1)$$ *(the numbers $\tan^2\theta_k$ are the roots of the reversed polynomial $x^nP_n(1/x)$)*, and verify it at $n  = 1$.

**Part IV — The Bertrand panorama.** For $\alpha,
\beta \in \R$, consider the *Bertrand series*

$$
\sum_{n \geq 3} \frac{1}{n^{\alpha}(\ln n)^{\beta}} .
$$

16. Show that for $\alpha > 1$ the series converges, whatever $\beta$ *(compare with $n^{-(1+\alpha)/2}$)* .
17. Show that for $\alpha < 1$ it diverges, whatever $\beta$ .
18. For $\alpha = 1$ : using the series–integral comparison ( [Theorem 6.6](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#thm-b2-comparison-seriesintegral) ) with $f(t) =  \frac{1}{t(\ln t)^\beta}$ , prove convergence iff $\beta  > 1$ .
19. Iterate the frontier: show $\sum\frac{1}{n\ln n\,\ln\ln  n}$ diverges while $\sum\frac{1}{n\ln n\,(\ln\ln n)^2}$ converges.
20. Two traps: determine the nature of $$\sum_n \frac{1}{n^{1 + 1/\ln n}}  \qquad\text{and}\qquad  \sum_n \frac{1}{n^{1 + 1/\ln\ln n}}$$ *(compute $n^{1/\ln n}$ exactly; compare $n^{1/\ln\ln n}$ with every power of $\ln n$)*.
21. (No slowest convergent series) Let $\sum a_n$ be any convergent series with $a_n > 0$ , and $R_n = \sum_{k  \geq n}a_k$ its tails. Prove that $\sum  \frac{a_n}{\sqrt{R_n}}$ still *converges* *(compare with the telescoping $2(\sqrt{R_n} -  \sqrt{R_{n+1}})$)* , although $\frac{a_n/\sqrt{R_n}}{a_n}  \to \infty$ : every convergent series is strictly dominated by another convergent series. The frontier of convergence is not a curve but a fog.

**Part V — Cross-checks and synthesis.**

22. (Cauchy condensation) Prove: for $(a_n)$ positive decreasing, $\sum a_n$ converges iff $\sum 2^k a_{2^k}$ converges. Re-derive question 18’s frontier from it.
23. (The cost of slowness) For $\sum\frac1{n(\ln n)^2}$ , bound the tail by an integral and show that summing up to $N = 10^6$ still leaves an error larger than $0.07$ : convergence certified by theory can be useless for numerics — contrast with question 13.
24. Classify (with one-line justifications): $\sum\frac1{n\ln n}$ , $\sum\frac1{n^{1.01}}$ , $\sum\frac{(\ln n)^{100}}{n^{1.001}}$ , $\sum\frac1{n(\ln n)(\ln\ln n)^{3}}$ .
25. (Synthesis) One sentence each: how de Moivre turned a trigonometric identity into a polynomial with computable root sums; where the squeeze needed *exact* endpoint identities rather than equivalents; which tool of [Chapter 6](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#ch-b2-comparison) powered Part IV; and what question 21 says about the dream of a “universal comparison test”. Name the two summits: Euler’s $\zeta(2) = \frac{\pi^2}6$ (and its floor above, $\zeta(4) = \frac{\pi^4}{90}$ ), and the Bertrand classification. Note where $\zeta(2)$ will be proved again: by Parseval in [Chapter 14](https://one-course.com/books/math/4/en/chapter/14-fourier-series#ch-b2-fourier) — one theorem, two civilizations.

**Solution of Problem 7.1.**

**1.** Induction on $m$: for $m = 0$ both sides are $1$; the step multiplies by $\cos\theta + \iu\sin\theta$ and uses the addition formulas $\cos(m\theta + \theta) =
\cos m\theta\cos\theta - \sin m\theta\sin\theta$, $\sin(m\theta
+ \theta) = \sin m\theta\cos\theta + \cos m\theta\sin\theta$. Expanding instead by the binomial theorem with $m = 2n+1$ and collecting the imaginary part (the odd powers of $\iu\sin\theta$, with $\iu^{2j+1} = (-1)^j\iu$):

$$
\sin\bigl((2n{+}1)\theta\bigr) = \sum_{j=0}^{n}(-1)^j
\binom{2n+1}{2j+1}\cos^{2(n-j)}\theta\,\sin^{2j+1}\theta .
$$

**2.** On $\intoo0{\frac\pi2}$, $\sin\theta \neq 0$: factor $\sin^{2n+1}\theta$ from each term, leaving $\bigl(\frac{\cos^2\theta}{\sin^2\theta}\bigr)^{n-j} =
(\cot^2\theta)^{n-j}$: the displayed identity with $P_n(x) =
\sum_j(-1)^j\binom{2n+1}{2j+1}x^{n-j}$. Its degree-$n$ coefficient is $j = 0$’s $\binom{2n+1}{1} = 2n + 1 \neq 0$.

**3.** At $\theta_k = \frac{k\pi}{2n+1}$: $\sin\bigl((2n{+}1)\theta_k\bigr) = \sin k\pi = 0$ while $\sin^{2n+1}\theta_k \neq 0$, so $P_n(\cot^2\theta_k) = 0$. The $\theta_k$ increase strictly in $\intoo0{\frac\pi2}$, where $\cot^2$ is strictly decreasing: the values $x_k =
\cot^2\theta_k$ are pairwise distinct — $n$ distinct roots of a degree-$n$ polynomial, hence all of them.

**4.** Vieta: the sum of the roots is minus the ratio of the $x^{n-1}$- and $x^n$-coefficients:

$$
\sum_{k=1}^{n}\cot^2\theta_k =
\frac{\binom{2n+1}{3}}{\binom{2n+1}{1}}
= \frac{(2n+1)(2n)(2n-1)/6}{2n+1} = \frac{n(2n-1)}{3}.
$$

**5.** $\frac{1}{\sin^2\theta} = 1 + \cot^2\theta$: summing, $n + \frac{n(2n-1)}3 = \frac{3n + 2n^2 - n}{3} =
\frac{2n(n+1)}{3}$.

**6.** On $\intoo{0}{\frac\pi2}$: $\sin\theta < \theta <
\tan\theta$ (Year 1 volume). Taking reciprocals reverses: $\cot\theta < \frac1\theta < \frac1{\sin\theta}$, and squaring (all positive) gives $\cot^2\theta < \frac1{\theta^2} <
\frac1{\sin^2\theta}$.

**7.** Sum question 6 at $\theta = \theta_k$ over $k \leq
n$, using questions 4 and 5, and $\frac1{\theta_k^2} =
\frac{(2n+1)^2}{k^2\pi^2}$:

$$
\frac{n(2n-1)}3 < \frac{(2n+1)^2}{\pi^2}\sum_{k=1}^n\frac1{k^2}
< \frac{2n(n+1)}3 .
$$

**8.** Multiply by $\frac{\pi^2}{(2n+1)^2}$:

$$
\frac{\pi^2}{3}\cdot\frac{n(2n-1)}{(2n+1)^2}
< \sum_{k=1}^{n}\frac1{k^2} <
\frac{\pi^2}{3}\cdot\frac{2n(n+1)}{(2n+1)^2}.
$$

Both bounds tend to $\frac{\pi^2}3\cdot\frac12 = \frac{\pi^2}6$ (the rational fractions tend to $\frac12$). The partial sums increase, so they converge, and the squeeze gives $\zeta(2) =
\frac{\pi^2}6$: Euler’s theorem, by Cauchy’s proof.

**9.** The partial sums increase to $\zeta(2) =
\frac{\pi^2}6$, so $0 \leq \frac{\pi^2}6 - \sum_{k\leq n}k^{-2}$; and the lower bound of question 8 gives

$$
\frac{\pi^2}6 - \sum_{k\leq n}\frac1{k^2}
\leq \frac{\pi^2}6 - \frac{\pi^2}3\cdot\frac{n(2n-1)}{(2n+1)^2}
= \frac{\pi^2}6\cdot\frac{(2n+1)^2 - (4n^2 -
2n)}{(2n+1)^2}
= \frac{\pi^2}6\cdot\frac{6n + 1}{(2n+1)^2}
= O\Bigl(\frac1n\Bigr),
$$

matching the exact tail $\frac1n - \frac1{2n^2} + O(n^{-3})$ of [Exercise 6.11](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#exo-b2-comparison-11).

**10.** The second elementary symmetric function of the roots is $\sigma_2 = \frac{\binom{2n+1}5}{\binom{2n+1}1} =
\frac{(2n)(2n-1)(2n-2)(2n-3)}{120}$, so

$$
\sum_k\cot^4\theta_k = \sigma_1^2 - 2\sigma_2
= \Bigl(\frac{n(2n-1)}3\Bigr)^2 -
\frac{2n(2n-1)(2n-2)(2n-3)}{60}
\sim \frac{4n^4}9 - \frac{4n^4}{15} = \frac{8n^4}{45}.
$$

Squeezing $\cot^4\theta < \theta^{-4} < (1 + \cot^2\theta)^2 =
1 + 2\cot^2\theta + \cot^4\theta$ and summing: both outer sums are $\frac{8n^4}{45}(1 + o(1))$ (the added $n + 2\sigma_1 =
O(n^2)$ is negligible), while the middle is $\frac{(2n+1)^4}{\pi^4}\sum_{k\leq n}k^{-4}$. Hence

$$
\sum_{k\leq n}\frac1{k^4} \longrightarrow
\pi^4\cdot\frac{8/45}{16} = \frac{\pi^4}{90}.
$$

**11.** Splitting $\zeta(2)$ over parities: $\sum_{\text{even}}
= \sum_j\frac{1}{(2j)^2} = \frac14\zeta(2) = \frac{\pi^2}{24}$, so $\sum_{\text{odd}} = \zeta(2) - \frac{\pi^2}{24} =
\frac{\pi^2}8$. Alternating: $\sum_k\frac{(-1)^{k-1}}{k^2} =
\sum_{\text{odd}} - \sum_{\text{even}} = \frac{\pi^2}8 -
\frac{\pi^2}{24} = \frac{\pi^2}{12}$ (absolute convergence justifies the regrouping, [Theorem 7.11](#thm-b2-series-rearrangement)).

**12.** $S = \frac{\zeta(2)^2}{\zeta(4)} =
\frac{(\pi^2/6)^2}{\pi^4/90} = \frac{90}{36} = \frac52$. Heuristic: the identity $\zeta(2)^2 = \zeta(4)S$ of [Exercise 7.7](#exo-b2-series-7) says that pulling out the $\gcd$ renormalizes pairs into coprime pairs; the reciprocal $\frac1{\zeta(2)} = \frac6{\pi^2} \approx 0.608$ is the natural candidate for the density of coprime pairs among all pairs — a statement about $\lim_N \frac{1}{N^2}\#\{(m,n) \leq N :
\gcd = 1\}$ whose honest proof (with error terms) belongs to the Year 3 volume.

**13.** Direct summation has error $\sim \frac1n$: six digits require about $10^6$ terms. The corrected sum $\sum_{k\leq n}k^{-2} + \frac1n - \frac1{2n^2}$ has error $O(n^{-3})$: at $n = 100$ it equals $1.6449339\dots$ against $\frac{\pi^2}6 = 1.6449341\dots$ — error $1.7\cdot10^{-7}$, seven digits from a hundred terms. Asymptotic corrections beat raw patience by four orders of magnitude.

**14.** $n = 1$: $P_1(x) = 3x - 1$, root $\frac13$, and indeed $\cot^2\frac\pi3 = \bigl(\frac1{\sqrt3}\bigr)^2 =
\frac13 = \frac{1\cdot1}3$. $n = 2$: the formula predicts $\frac{2\cdot3}3 = 2$; with $\cos\frac\pi5 = \frac{1 +
\sqrt5}{4}$, one computes $\cot^2 36^\circ \approx 1.894$ and $\cot^2 72^\circ \approx 0.106$: sum $2.000$.

**15.** The numbers $\tan^2\theta_k = \frac1{x_k}$ are the roots of $Q(x) = x^nP_n\bigl(\frac1x\bigr) =
\sum_{j=0}^n(-1)^j\binom{2n+1}{2j+1}x^j$ (the $x_k$ are nonzero). Vieta on $Q$: the leading coefficient is $(-1)^n$ (term $j = n$), the next is $(-1)^{n-1}\binom{2n+1}{2n-1} =
(-1)^{n-1}\binom{2n+1}{2}$, so

$$
\sum_{k=1}^n\tan^2\theta_k =
-\frac{(-1)^{n-1}\binom{2n+1}2}{(-1)^n} = \binom{2n+1}2\cdot
\frac{2}{2n+1}\cdot\frac{2n+1}{2}
= n(2n+1).
$$

Check $n = 1$: $\tan^2\frac\pi3 = 3 = 1\cdot3$.

**16.** Let $\gamma = \frac{1+\alpha}2 \in
\intoo{1}{\alpha}$. Then $\frac{n^{-\alpha}(\ln
n)^{-\beta}}{n^{-\gamma}} = n^{\gamma - \alpha}(\ln n)^{-\beta}
\to 0$ (a negative power of $n$ beats any power of $\ln n$), so eventually the terms are $\leq n^{-\gamma}$ with $\gamma > 1$: convergence by comparison with a Riemann series.

**17.** Let $\gamma = \frac{1+\alpha}2 \in
\intoo{\alpha}{1}$: now $\frac{n^{-\gamma}}{n^{-\alpha}(\ln
n)^{-\beta}} = n^{\alpha-\gamma}(\ln n)^{\beta} \to 0$, so eventually the terms are $\geq n^{-\gamma}$ with $\gamma < 1$: divergence.

**18.** $f(t) = \frac1{t(\ln t)^\beta}$ is positive, [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity), and decreasing for large $t$ (its logarithm has derivative $-\frac1t\bigl(1 + \frac{\beta}{\ln t}\bigr) < 0$ eventually). Antiderivatives: for $\beta \neq 1$, $\int^x f = \frac{(\ln x)^{1-\beta}}{1-\beta} + \text{const}$, which has a finite limit iff $\beta > 1$; for $\beta = 1$, $\int^x f = \ln\ln x \to \infty$. By [Theorem 6.6](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#thm-b2-comparison-seriesintegral), the series and the integral share their nature: convergence iff $\beta > 1$.

**19.** Same test: $\frac{\dd}{\dd t}\ln\ln\ln t =
\frac{1}{t\ln t\,\ln\ln t}$, and $\ln\ln\ln t \to \infty$: divergence. And $\frac{\dd}{\dd t}\Bigl(-\frac1{\ln\ln t}\Bigr)
= \frac{1}{t\ln t\,(\ln\ln t)^2}$ with $-\frac1{\ln\ln t} \to
0$: convergence.

**20.** First: $n^{1/\ln n} = \eu^{\ln n/\ln n} = \eu$, so the terms are exactly $\frac{1}{\eu\,n}$: a multiple of the harmonic series, *divergent* — the exponent $1 +
\frac1{\ln n}$ crawls to $1$ too fast. Second: $n^{1/\ln\ln n}
= \eu^{\ln n/\ln\ln n}$, and $\frac{\ln n}{\ln\ln n} \geq
2\ln\ln n$ eventually, so $n^{1/\ln\ln n} \geq (\ln n)^2$: the terms are $\leq \frac1{n(\ln n)^2}$, a convergent Bertrand series (question 18): *convergent*. The frontier passes strictly between these two exponents.

**21.** $R_n \downarrow 0$ and

$$
\sqrt{R_n} - \sqrt{R_{n+1}} = \frac{R_n -
R_{n+1}}{\sqrt{R_n} + \sqrt{R_{n+1}}} =
\frac{a_n}{\sqrt{R_n} + \sqrt{R_{n+1}}} \geq
\frac{a_n}{2\sqrt{R_n}},
$$

so $\sum_n \frac{a_n}{\sqrt{R_n}} \leq 2\sum_n(\sqrt{R_n} -
\sqrt{R_{n+1}}) = 2\sqrt{R_1} < \infty$ (telescoping). Yet $\frac{a_n/\sqrt{R_n}}{a_n} = \frac1{\sqrt{R_n}} \to \infty$: the new series converges while being infinitely larger. No convergent series is slowest; comparison tests against any fixed family can never be complete.

**22.** For decreasing positive $(a_n)$, group the terms between consecutive powers of $2$:

$$
2^{k}a_{2^{k+1}} \leq \sum_{n=2^k}^{2^{k+1}-1} a_n \leq
2^ka_{2^k} .
$$

Summing over $k$: if $\sum 2^ka_{2^k}$ converges, the partial sums of $\sum a_n$ are bounded (converges); if $\sum a_n$ converges, then $\sum_k 2^{k+1}a_{2^{k+1}} \leq 2\sum_n a_n <
\infty$. For $a_n = \frac1{n(\ln n)^\beta}$: $2^ka_{2^k} =
\frac{1}{(k\ln 2)^\beta}$, and $\sum k^{-\beta}$ converges iff $\beta > 1$: the frontier of question 18 again, without integrals.

**23.** By the integral comparison,

$$
\sum_{n > N}\frac{1}{n(\ln n)^2} \geq
\int_{N+1}^{\infty}\frac{\dd t}{t(\ln t)^2} =
\frac{1}{\ln(N+1)},
$$

which at $N = 10^6$ is $\approx 0.0724$: after a million terms the tail still exceeds $0.07$ — the series converges, but no direct summation will ever exhibit its sum. Contrast with question 13, where one asymptotic correction bought seven digits from a hundred terms: knowing *how* a series converges is worth more than knowing that it does.

**24.** $\sum\frac1{n\ln n}$: diverges ($\alpha = 1$, $\beta = 1$, question 18). $\sum\frac1{n^{1.01}}$: converges (Riemann, $\alpha > 1$). $\sum\frac{(\ln n)^{100}}{n^{1.001}}$: converges ($\alpha = 1.001 > 1$, $\beta = -100$, question 16). $\sum\frac1{n\ln n(\ln\ln n)^3}$: converges (question 19’s pattern: antiderivative $-\frac12(\ln\ln t)^{-2}$, finite limit).

**25.** De Moivre converts the vanishing of $\sin(2n{+}1)
\theta_k$ into the vanishing of a polynomial at $\cot^2\theta_k$, and Vieta reads off the exact root sums that analysis alone could only estimate (questions 1–5). The squeeze needed the *exact* values $\frac{n(2n-1)}3$ and $\frac{2n(n+1)}3$ on both sides — equivalents would have begged the question, since the whole point is the constant $\frac{\pi^2}6$ (questions 7–8). Part IV ran entirely on [Chapter 6](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#ch-b2-comparison)’s series–integral comparison, the logarithmic antiderivatives doing the classifying (questions 18–19). Question 21 destroys the dream of a universal comparison test: below every convergent series lies another, infinitely slower — scales like Bertrand’s map the frontier ever more finely but never reach it. Summits: Euler’s $\zeta(2)
= \frac{\pi^2}6$ with its upper floor $\zeta(4) =
\frac{\pi^4}{90}$ (questions 8, 10), and the Bertrand classification (questions 16–18); $\zeta(2)$ returns in [Chapter 14](https://one-course.com/books/math/4/en/chapter/14-fourier-series#ch-b2-fourier), where Parseval’s identity re-proves it in one line from the Fourier series of the sawtooth — one constant, two civilizations.