University Mathematics — Year 2 · Bachelor Year 2

7Sequences and Series

The theory of numerical series (Year 1 volume) matures here in three directions: series with values in Banach spaces, where absolute convergence does the work; the finer tests for real series (Abel summation); and summable families — summation freed from the order of the terms — with the Fubini theorem for double sums and the Cauchy product. 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); in a non-complete space this may fail (Exercise 7.9).

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 to $(I - A)^{-1}$ (proved in Exercise 5.5); $\sum \frac{A^k}{k!}$ converges absolutely to $\eu^A$ for every $A$ (Example 5.22). 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), 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, 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 ( $\abs{\sin n} \geq \sin^2 n = \frac{1 - \cos 2n}{2}$ , 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 ( $\sum\frac1n$ ). 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), 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.

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 — 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 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 is the canonical counterexample, and Exercise 7.1’s third series the drill. (iii) Conditionally convergent series may not be rearranged (Example 7.12), and their Cauchy products 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) is not a luxury. (iv) Summability is about absolute bounds by definition: there is no such thing as a conditionally summable family.

7.3 Summable families

Definition 7.8

Let $I$ be a 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) 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 convergent or a clean test case (Example 7.6). (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 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? 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 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. (The same census shows $\bigl((m^2 + n^2)^{-1}\bigr)$ is not 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)

For nonnegative families, the sum is invariant under any enumeration: $\sum_{i} u_i = \sum_{n=0}^{\infty} u_{\sigma(n)}$ for every bijection $\sigma \colon \N \to I$ .
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 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 — 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 convergent series remember the order of their terms; 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 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 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). In particular, if $\sum a_m$ and $\sum b_n$ converge absolutely, their Cauchy product converges absolutely 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 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. The family $u_{m,n} = a_m b_n$ is 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.)

Example 7.16 (Cauchy products as a computing device)

From the geometric series and Exercise 7.2’s $\sum_{n \geq 1} nz^n = \frac{z}{(1-z)^2}$ ( $\abs z < 1$ ), one more Cauchy product 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 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).

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 (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 families turn combinatorics into analysis.

Example 7.18 (A Fubini evaluation: $\sum_n (\zeta(n) - 1) = 1$ )

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: 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 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 $\bigl(C(\intcc01), \norm\cdot_\infty\bigr)$ , solve $x - K(x) = y$ where $K(f)$ is the constant function $\frac12\int_0^1 f$ . The operator norm is $\vertiii K \leq \frac12 < 1$ , so the Neumann series applies (Example 7.2): $x = \sum_{n\geq0} K^n(y)$ . 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 (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: normal convergence of $\sum f_n$ is absolute convergence of $\sum\norm{f_n}_\infty$ in the Banach space $\bigl(C, \norm\cdot_\infty\bigr)$ — this chapter’s Theorem 5.21 in costume. For Chapter 11: inside the disk of convergence everything is absolute and summable, so Cauchy products and rearrangements run free (that is why power series multiply like polynomials); on the boundary, Abel’s test takes over (Example 7.5). For Chapter 23: 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. 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) are summable families in disguise, Fourier coefficients get multiplied by Cauchy products and rearranged by Parseval (Chapter 14), and probability generating functions (Chapter 23) are Fubini’s theorem applied to expectations. The Year 3 volume absorbs summable families into Lebesgue integration over the counting measure — where Theorem 7.14 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

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): convergent (not absolutely, 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 ( $\ln n \leq n$ ).

$\sum \frac{(-1)^n}{n^{3/4} + \cos n}$ : 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 convergent ( $\frac{1}{n^{3/2}}$ scale). Third: absolutely 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 of $\sum z^n$ with itself.

Solution

Solution of Exercise 7.2.

Cauchy product of $\sum_{m\geq0} z^m$ with itself (both absolutely 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, that $\dfrac{1}{n}\sum_{k=1}^{n} k\,b_k \to 0$ .

Solution

Solution of Exercise 7.3.

Let $B_n = \sum_{k \leq n} b_k \to B$ . Abel summation 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 convergent, semi-convergent, divergent?

Solution

Solution of Exercise 7.4.

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

$\alpha > 1$ : absolutely 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: $\abs{\sin n\theta} \geq \sin^2 n\theta = \frac{1 - \cos 2n\theta}{2}$ , 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 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

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 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

Solution of Exercise 7.6.

Summability: 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):

(\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, 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

Solution of Exercise 7.7.

Summability: bounded by $\zeta(2)^2$ as a product family (Theorem 7.14’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 accordingly (a partition of the index set — legitimate for summable families by Theorem 7.11/Theorem 7.14 applied to the partition into countably 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: $S = \frac{5}{2}$ .)

Exercise 7.8 ★★★

(Abel’s theorem on products, light version) Suppose $\sum a_n$ converges absolutely and $\sum b_n$ converges. Prove that their Cauchy product $\sum c_n$ 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

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) space $E$ of eventually-zero real sequences with the sup norm, exhibit an absolutely convergent series that does not converge in $E$ . (Try $u_n = 2^{-n} e_n$ with $(e_n)$ the canonical sequences.)

Solution

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 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 is exactly what Theorem 5.21 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

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

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

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, 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.

Prove de Moivre’s formula $(\cos\theta + \iu\sin\theta)^m = \cos m\theta + \iu\sin m\theta$ ( $m \in \N$ ), and deduce, for $m = 2n + 1$ ,

$\sin\bigl((2n{+}1)\theta\bigr) = \sum_{j=0}^{n} (-1)^j\binom{2n+1}{2j+1}\cos^{2(n-j)}\theta\, \sin^{2j+1}\theta .$
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 $2n + 1$ .
Show that $x_k = \cot^2\theta_k$ , $k = 1, \dots, n$ , are $n$ distinct roots of $P_n$ — hence all of them.
By Vieta, prove the exact identity

$\sum_{k=1}^{n} \cot^2\frac{k\pi}{2n+1} = \frac{n(2n-1)}{3}.$
Deduce also $\displaystyle\sum_{k=1}^{n} \frac{1}{\sin^2\theta_k} = \frac{2n(n+1)}{3}$ .
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.

Summing question 6 over $k = 1, \dots, n$ with $\theta = \theta_k$ , establish

$\frac{n(2n-1)}{3} \;<\; \frac{(2n+1)^2}{\pi^2}\sum_{k=1}^{n}\frac1{k^2} \;<\; \frac{2n(n+1)}{3}.$
Conclude (Euler’s theorem, by Cauchy’s proof):

$\zeta(2) = \sum_{k=1}^{\infty}\frac{1}{k^2} = \frac{\pi^2}{6}.$
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.
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.

Deduce from $\zeta(2) = \frac{\pi^2}6$ :

$\sum_{k\geq0}\frac{1}{(2k+1)^2} = \frac{\pi^2}{8}, \qquad \sum_{k\geq1}\frac{(-1)^{k-1}}{k^2} = \frac{\pi^2}{12}.$
Combine with Exercise 7.7: 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).
(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 $1.7\cdot10^{-7}$ ).
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.
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}} .

Show that for $\alpha > 1$ the series converges, whatever $\beta$ (compare with $n^{-(1+\alpha)/2}$ ).
Show that for $\alpha < 1$ it diverges, whatever $\beta$ .
For $\alpha = 1$ : using the series–integral comparison (Theorem 6.6) with $f(t) = \frac{1}{t(\ln t)^\beta}$ , prove convergence iff $\beta > 1$ .
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.
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 $\ln n$ ).
(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.

(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.
(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.
Classify (with one-line justifications): $\sum\frac1{n\ln n}$ , $\sum\frac1{n^{1.01}}$ , $\sum\frac{(\ln n)^{100}}{n^{1.001}}$ , $\sum\frac1{n(\ln n)(\ln\ln n)^{3}}$ .
(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 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 — one theorem, two civilizations.

Solution

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.

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).

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 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, 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, 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’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, where Parseval’s identity re-proves it in one line from the Fourier series of the sawtooth — one constant, two civilizations.

7.1 Series in normed spaces

7.2 Abel summation

7.3 Summable families

7.4 Exercises

7.5 Problem: Euler’s ζ(2)=π2/6\zeta(2) = \pi^2/6ζ(2)=π2/6, by Cauchy’s Cotangent Sum

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