University Mathematics — Year 1 · Bachelor Year 1

17Numerical Series

Summing infinitely many numbers means taking the limit of the partial sums — nothing more, nothing less. This chapter sets up the definitions and the convergence tests usable in first year: comparison and equivalents for positive terms, the ratio test, the integral comparison giving the Riemann series, absolute convergence, and the alternating series theorem. The finer theory (products of series, summation by packets, series of functions) belongs to the second year.

17.1 Generalities

Definition 17.1

Given a sequence $(u_n)$ , the series $\sum u_n$ is the sequence of partial sums $S_N = \sum_{n=0}^{N} u_n$ . The series converges when $(S_N)$ converges; the limit is the sum $\sum_{n=0}^{\infty} u_n$ , and $R_N = \sum_{n > N} u_n = S - S_N$ is the remainder, which tends to $0$ .

Example 17.2 (Geometric series)

For $q \in \C$ : $\;S_N = \sum_{n=0}^{N} q^n = \frac{1 - q^{N+1}}{1-q}$ ( $q \neq 1$ ). The series converges iff $\abs q < 1$ (Exercise 11.3), with

\sum_{n=0}^{\infty} q^n = \frac{1}{1 - q} .

Example 17.3 (Periodic decimals are geometric series)

What number is $0.363636\dots$ ? Its very writing is a series:

0.\overline{36} = \sum_{k=1}^{\infty} \frac{36}{100^k} = 36\cdot\frac{1/100}{1 - 1/100} = \frac{36}{99} = \frac{4}{11} ,

by the geometric sum with $q = \frac{1}{100}$ . In general a block $B$ of $p$ digits repeating forever is worth $\frac{B}{10^p - 1}$ — the mechanism behind the periodicity criterion of Problem 10.1, which this chapter’s language finally states in one line: a decimal expansion is a convergent series, eventually periodic exactly when its sum is rational. The digit machinery of Chapter 10, built there with bare suprema, was series theory traveling incognito.

Proposition 17.4 (First facts)

If $\sum u_n$ converges, then $u_n \to 0$ . (The converse is false: the harmonic series.)
Linearity: convergent series add and scale, with the expected sums.
(Telescoping) $\sum (v_{n+1} - v_n)$ converges iff $(v_n)$ converges, with sum $\lim v_n - v_0$ .
Changing finitely many terms does not affect convergence (only the sum).

Proof. (1) $u_N = S_N - S_{N-1} \to S - S = 0$ . The harmonic series has $u_n = \frac1n \to 0$ yet diverges (Exercise 11.5). (2) Operations on limits. (3) $S_N = v_{N+1} - v_0$ . (4) The partial sums change by an eventually constant amount. ∎

Example 17.5 (Planning digits with the geometric remainder)

For $\abs q < 1$ the remainder of the geometric series is explicit:

R_N = \sum_{n = N+1}^{\infty} q^n = \frac{q^{N+1}}{1 - q} .

This converts accuracy goals into term counts before any computation. To evaluate $\sum_{n\geq0} \bigl(\frac13\bigr)^n = \frac32$ within $10^{-10}$ : need $\frac{(1/3)^{N+1}}{2/3} \leq 10^{-10}$ , i.e. $3^{N} \geq \frac{3}{2}\cdot 10^{10}$ , i.e. $N \geq 22$ (as $3^{22} \approx 3.1\cdot10^{10}$ ): twenty-three terms, known in advance. Every geometric-rate estimate of the weekend problems (the $\frac13$ -series for $\ln 2$ , Machin’s arctangents in Problem 16.1) is this two-line budget in professional dress.

Example 17.6 (A longer telescope)

Compute $\sum_{n\geq1} \frac{1}{n(n+1)(n+2)}$ . Partial fractions (Chapter 9):

\frac{1}{n(n+1)(n+2)} = \frac{1/2}{n} - \frac{1}{n+1} + \frac{1/2}{n+2} = \frac12\Bigl(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\Bigr),

where the second form — a difference of consecutive values of $w_n = \frac{1}{n(n+1)}$ — is the telescoping one. Hence

\sum_{n=1}^{N} \frac{1}{n(n+1)(n+2)} = \frac12\Bigl(w_1 - w_{N+1}\Bigr) = \frac12\Bigl(\frac12 - \frac{1}{(N+1)(N+2)}\Bigr) \longrightarrow \frac14 .

The closing insight: three-term partial fractions rarely telescope as written; regroup them into a difference $w_n - w_{n+1}$ first — the reward is not just convergence but the exact sum, which no comparison test ever delivers.

17.2 Series with nonnegative terms

Theorem 17.7 (Bounded partial sums)

If $u_n \geq 0$ for all $n$ , the partial sums increase, so: $\sum u_n$ converges $\iff$ its partial sums are bounded above. Hence the comparison test: if $0 \leq u_n \leq v_n$ for all (large) $n$ ,

\sum v_n \text{ converges} \implies \sum u_n \text{ converges}, \qquad \sum u_n \text{ diverges} \implies \sum v_n \text{ diverges}.

And the equivalents test: if $u_n \sim v_n$ with $v_n \geq 0$ , the two series have the same nature.

Proof. Monotone limit theorem (Theorem 11.9) for the first point; comparison of partial sums for the second. Equivalents: for large $n$ , $\frac12 v_n \leq u_n \leq 2 v_n$ (definition of $\sim$ with $\varepsilon = \frac12$ ), and comparison applies both ways. ∎

Example 17.8 (An equivalent that proves divergence)

Nature of $\sum_{n\geq1} n\sin\dfrac{1}{n^2}$ ? Since $\frac{1}{n^2} \to 0$ and $\sin h \sim h$ at $0$ :

n\sin\frac{1}{n^2} \;\sim\; n\cdot\frac{1}{n^2} = \frac1n ,

and the equivalents test transfers the divergence of the harmonic series: divergent — even though the terms tend to $0$ . One expansion, one scale, one verdict; the same two-step pattern (equivalent, then Riemann or geometric lookup) decides all four series of Exercise 17.3.

Example 17.9 (The equivalents test in one line)

Nature of $\sum_{n \geq 1} \frac{\sqrt{n+1} - \sqrt n}{n}$ ? Conjugate the numerator:

\frac{\sqrt{n+1} - \sqrt n}{n} = \frac{1}{n\,(\sqrt{n+1} + \sqrt n)} \sim \frac{1}{2\,n^{3/2}} ,

a convergent Riemann scale ( $\alpha = \frac32 > 1$ ): the series converges. The whole decision took one equivalent and one lookup — provided the terms are nonnegative, which they are. The closing insight: for positive series, the entire convergence theory is a dictionary of scales ( $n^{-\alpha}$ , $q^n$ , $\frac{1}{n(\ln n)^\alpha}$ ) plus the license to replace a term by an equivalent; the analytic work is in the asymptotics (Chapter 16), never in the summation.

Theorem 17.10 (Integral comparison; Riemann series)

Let $f$ be continuous, nonnegative and decreasing on $\intco{1}{+\infty}$ . Then

\int_1^{N+1} f(t)\,\dd t \;\leq\; \sum_{n=1}^{N} f(n) \;\leq\; f(1) + \int_1^{N} f(t)\,\dd t ,

so $\sum f(n)$ converges iff $\bigl(\int_1^x f\bigr)$ is bounded. In particular, for $\alpha \in \R$ :

\sum_{n \geq 1} \frac{1}{n^\alpha} \text{ converges} \iff \alpha > 1,

and $\sum_{n=1}^{N} \frac1n = \ln N + O(1)$ .

Proof. For $n \leq t \leq n+1$ , monotonicity gives $f(n+1) \leq f(t) \leq f(n)$ ; integrating over $\intcc{n}{n+1}$ (a segment of length $1$ ):

f(n+1) \;\leq\; \int_n^{n+1} f(t)\,\dd t \;\leq\; f(n) .

Summing the right inequalities for $n = 1, \dots, N-1$ gives $\int_1^{N} f \leq \sum_{n=1}^{N-1} f(n)$ , hence the upper framing after adding $f(N) \leq f(1)$ ; summing the left ones for $n = 1, \dots, N$ gives $\sum_{n=2}^{N+1} f(n) \leq \int_1^{N+1} f$ , which after reindexing is the lower framing. Convergence: the partial sums and the integrals $\int_1^x f$ bound each other within the constant $f(1)$ , and both are nondecreasing, so one is bounded iff the other is (Theorem 17.7). For $f(t) = t^{-\alpha}$ ( $\alpha \neq 1$ ): $\int_1^x t^{-\alpha}\dd t = \frac{x^{1-\alpha} - 1}{1 - \alpha}$ , bounded iff $\alpha > 1$ ; for $\alpha = 1$ the integral is $\ln x \to \infty$ , and the framing gives $\ln(N+1) \leq H_N \leq 1 + \ln N$ . For $\alpha \leq 0$ the terms do not tend to $0$ . ∎

Example 17.11 (The harmonic stack)

How many terms must the harmonic series accumulate to pass $20$ ? The framing $\ln(N+1) \leq H_N \leq 1 + \ln N$ answers with no summation at all: $H_N \geq 20$ requires $1 + \ln N \geq 20$ , i.e. $N \geq \eu^{19} \approx 1.8\cdot10^{8}$ , and is guaranteed once $\ln(N + 1) \geq 20$ , i.e. $N \approx \eu^{20} \approx 4.9\cdot10^{8}$ . (The weekend problem sharpens this to $N \approx \eu^{20 - \gamma} \approx 2.7\cdot10^{8}$ via Euler’s constant.) The closing insight: the integral comparison does not merely decide convergence — it locates partial sums with logarithmic precision, turning a hopeless computation (hundreds of millions of terms) into a two-line estimate.

Theorem 17.12 (Ratio test (d’Alembert))

Let $u_n > 0$ with $\frac{u_{n+1}}{u_n} \to \ell$ .

If $\ell < 1$ : $\sum u_n$ converges;
if $\ell > 1$ : $u_n \to +\infty$ , divergence;
if $\ell = 1$ : no conclusion ( $\sum \frac1n$ diverges, $\sum \frac{1}{n^2}$ converges).

Proof. If $\ell < 1$ , fix $q \in \intoo{\ell}{1}$ : beyond some $N$ , $u_{n+1} \leq q\,u_n$ , so $u_n \leq u_N q^{\,n-N}$ by induction: comparison with a geometric series. If $\ell > 1$ : beyond some $N$ the sequence $(u_n)$ is increasing, so it cannot tend to $0$ (its limit, if any, is $\geq u_N > 0$ ); by Proposition 17.4 (1), divergence — and in fact $u_n \geq u_N q^{n-N}$ with $q > 1$ gives $u_n \to \infty$ . ∎

Example 17.13

$\sum \frac{x^n}{n!}$ converges for every $x > 0$ : ratio $\frac{x}{n+1} \to 0$ . Its sum is $\eu^x$ : by Taylor–Lagrange (Theorem 16.7) on $\intcc{0}{x}$ ,

\Bigl| \eu^x - \sum_{k=0}^{n} \frac{x^k}{k!} \Bigr| \leq \eu^{x}\, \frac{x^{n+1}}{(n+1)!} \xrightarrow[n\to\infty]{} 0 ,

the bound tending to $0$ because the factorial dominates (Exercise 15.9 (1) used the same fact). The same argument sums the $\sin$ , $\cos$ , $\sinh$ , $\cosh$ series on all of $\R$ .

Example 17.14 (The ratio test is sufficient, not necessary)

Let $u_n = 2^{-n}$ for even $n$ and $u_n = 2^{-n-2}$ for odd $n$ . The consecutive ratios oscillate between $\frac{1}{8}$ and $\frac12\cdot4 = 2$ , so $\frac{u_{n+1}}{u_n}$ has no limit and d’Alembert is mute — yet $u_n \leq 2^{-n}$ and the comparison test settles convergence instantly. The test’s hypothesis (the ratio converges) is a real restriction: it suits terms with one dominant multiplicative structure (factorials, powers), and fails on anything that breathes. When ratios misbehave, step back to comparison against a geometric envelope — which is all the ratio test ever was, as its proof shows.

Example 17.15 (Ratio test on factorial battles)

Nature of $\sum_{n\geq0} \dfrac{(n!)^2}{(2n)!}$ (reciprocals of the central binomial coefficients, up to the factor $n + 1$ )? The ratio collapses the factorials:

\frac{u_{n+1}}{u_n} = \frac{((n+1)!)^2}{(n!)^2}\cdot\frac{(2n)!}{(2n+2)!} = \frac{(n+1)^2}{(2n+1)(2n+2)} \longrightarrow \frac14 < 1 :

convergent, with room to spare — the terms decay essentially like $4^{-n}$ , consistent with $\binom{2n}{n} \geq \frac{4^n}{2n+1}$ from Problem 15.1. The closing insight: quotients of factorials are exactly what the ratio test digests — every factorial cancels into a rational function of $n$ , whose limit is read off the leading terms.

17.3 Absolute convergence; alternating series

Theorem 17.16 (Absolute convergence)

If $\sum \abs{u_n}$ converges (absolute convergence), then $\sum u_n$ converges, and $\bigl|\sum u_n\bigr| \leq \sum \abs{u_n}$ . This holds for real or complex terms.

Proof. The partial sums satisfy, for $M > N$ (Cauchy criterion, Theorem 11.20):

\abs{S_M - S_N} = \Bigl| \sum_{n=N+1}^{M} u_n \Bigr| \leq \sum_{n=N+1}^{M} \abs{u_n},

which is small for large $N$ since the partial sums of $\sum\abs{u_n}$ form a Cauchy sequence. So $(S_N)$ is Cauchy, hence convergent. The inequality passes to the limit from the finite triangle inequality. ∎

Example 17.17 (Absolute convergence, real and complex)

$\sum_{n\geq1} \frac{\sin n}{n^2}$ : the terms change sign erratically (indeed $(\sin n)$ is dense in $\intcc{-1}{1}$ , Exercise 11.12), and no alternating structure is in sight. Absolute convergence rescues everything at once: $\bigl|\frac{\sin n}{n^2}\bigr| \leq \frac{1}{n^2}$ , a convergent scale, so the series converges. The same shield works over $\C$ : $\sum_{n\geq1}\frac{\eu^{\iu n}}{n^2}$ converges because $\bigl|\frac{\eu^{\iu n}}{n^2}\bigr| = \frac{1}{n^2}$ — sign patterns, even two-dimensional ones, are irrelevant once the moduli are summable. The closing insight: absolute convergence is the only tool of this chapter that never asks how the signs are organized; try it first (Method 17.21), and reserve the delicate tests for the series that fail it.

Theorem 17.18 (Alternating series test)

Let $(a_n)$ be decreasing with $a_n \to 0$ . Then the alternating series $\sum (-1)^n a_n$ converges; its sum lies between any two consecutive partial sums, and

\abs{R_N} = \Bigl| \sum_{n > N} (-1)^n a_n \Bigr| \leq a_{N+1} .

Proof. The even and odd partial sums are adjacent: $S_{2p+2} - S_{2p} = a_{2p+2} - a_{2p+1} \leq 0$ (decreasing), $S_{2p+1} - S_{2p-1} = a_{2p} - a_{2p+1} \geq 0$ (increasing), and $S_{2p} - S_{2p+1} = a_{2p+1} \to 0$ . By Theorem 11.11 they share a limit $S$ , which the two subsequences criterion (Proposition 11.14) makes the limit of $(S_N)$ ; moreover $S$ is trapped between consecutive partial sums, and $\abs{S - S_N}$ is at most the gap to the next one, $a_{N+1}$ . ∎

Example 17.19 (Alternating harmonic series)

$\sum_{n \geq 1} \frac{(-1)^{n-1}}{n}$ converges (alternating test) but not absolutely (harmonic series). Its sum is $\ln 2$ : from the finite geometric identity $\frac{1}{1+t} = \sum_{k=0}^{n-1} (-t)^k + \frac{(-t)^n}{1+t}$ , integrate over $\intcc{0}{1}$ :

\ln 2 = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} + (-1)^n \int_0^1 \frac{t^n}{1+t}\,\dd t , \qquad 0 \leq \int_0^1 \frac{t^n}{1+t}\,\dd t \leq \frac{1}{n+1} \to 0 .

The convergence is painfully slow ( $R_N \approx \frac{1}{N}$ ) — alternating series converge by cancellation, not by smallness.

The partial sums S_N of the alternating harmonic series 1 - 1/2 + 1/3 - hop over their limit 2 at every step: odd sums from above, even sums from below, each hop of size 1/N+1. The bracketing is the proof of made visible — and the slow closing of the pincers (|S_N - 2| 1/2N, weekend problem ) is why nobody computes 2 this way. — The partial sums $S_N$ of the alternating harmonic series $1 - \frac12 + \frac13 - \cdots$ hop over their limit $\ln 2$ at every step: odd sums from above, even sums from below, each hop of size $\frac{1}{N+1}$ . The bracketing is the proof of Theorem 17.18 made visible — and the slow closing of the pincers ( $\abs{S_N - \ln 2} \approx \frac{1}{2N}$ , weekend problem Problem 17.1) is why nobody computes $\ln 2$ this way.

Remark 17.20 (Common pitfalls with series)

(i) The equivalents test needs a sign: let $v_n = \frac{(-1)^n}{\sqrt n}$ and $u_n = v_n + \frac1n$ . Then $\frac{u_n}{v_n} = 1 + \frac{(-1)^n}{\sqrt n} \to 1$ , so $u_n \sim v_n$ ; yet $\sum v_n$ converges (alternating test) while $\sum u_n = \sum v_n + \sum \frac1n$ diverges. Equivalence controls the size of terms, and for signed series size is not destiny — the test is stated, and true, for (eventually) nonnegative terms only. (ii) $u_n \to 0$ proves nothing: the harmonic series is the eternal counterexample; the converse direction (Proposition 17.4 (1)) is only a quick divergence test. (iii) Ratio limit $1$ is silence, not convergence: both $\sum\frac1n$ and $\sum\frac{1}{n^2}$ have ratio $\to 1$ ; switch to Riemann scales or integral comparison. (iv) Alternating needs decreasing: $\sum \frac{(-1)^n}{n + (-1)^n}$ looks alternating and is handled only by expansion (Exercise 17.5); the weekend problem of Chapter 16 (question 23 there) shows the test can fail outright without monotonicity. (v) Grouping and reordering are not free: inserting parentheses is harmless for convergent series but can create convergence from divergence ( $1 - 1 + 1 - \cdots$ grouped in pairs), and reordering can change the sum itself — the drama staged in this chapter’s weekend problem (Problem 17.1).

Method 17.21 (Deciding the nature of a series)

Does $u_n \to 0$ ? If not, divergence, stop.
Nonnegative terms: seek an equivalent of $u_n$ (expansions, Chapter 16!), compare with Riemann or geometric scales; factorials and powers call for the ratio test; decreasing $f(n)$ calls for integral comparison.
Signs vary: try absolute convergence first; if it fails, the alternating test (check decreasing carefully); beyond that, second-year tools.

Example 17.22 (Odd denominators, half the telescope)

Compute $\sum_{n \geq 1} \dfrac{1}{4n^2 - 1}$ . Partial fractions: $\frac{1}{(2n-1)(2n+1)} = \frac12\bigl(\frac{1}{2n-1} - \frac{1}{2n+1}\bigr)$ , so

\sum_{n=1}^{N} \frac{1}{4n^2 - 1} = \frac12\Bigl(1 - \frac{1}{2N+1}\Bigr) \longrightarrow \frac12 .

Compare with $\sum \frac{1}{n(n+1)} = 1$ (Exercise 17.1): same telescoping skeleton, but the consecutive terms here are two apart in the odd numbers, and the factor $\frac12$ records the step. The closing insight: telescoping is a change of viewpoint, not a trick — whenever the general term is a difference $w_n - w_{n+1}$ of a sequence with a limit, the sum is $w_1 - \lim w$ , exactly Proposition 17.4 (3).

Remark 17.23 (The analysis pipeline, in retrospect)

This chapter is where the volume’s analysis converges, and each test names its ancestor. Bounded partial sums is the monotone limit theorem (Chapter 11), itself the completeness axiom of Chapter 10; absolute convergence is the Cauchy criterion; the integral test is Chapter 15’s framing of areas; equivalents of general terms are Chapter 16’s expansions; and the alternating theorem is the adjacent-sequences lemma in its Sunday clothes. Read backwards, the pipeline explains what each chapter was for — and the weekend problems threaded through it ( $b$ -adic digits, Cesàro–Stolz, the irrationality machines, Euler’s constant) are the same few ideas meeting at higher and higher altitude. The linear algebra that follows changes subject, not standards: the habit of exact statements with certified error survives the move from limits to dimensions.

Remark 17.24 (Where series go next)

This chapter closes the analysis of the volume and opens three doors. In the Year 2 volume, series acquire a variable ( $\sum a_n x^n$ : power series, with their radius of convergence) and then a function-valued theory (Fourier series); the dichotomy absolute-versus-conditional convergence, dramatized in the weekend problem below, becomes the cornerstone of both. In probability (Year 3 volume), expectations of discrete random variables are series, and absolute convergence is what makes them well defined. And the Riemann series $\sum n^{-s}$ , pushed to complex $s$ , becomes the zeta function — the single most studied series in mathematics.

17.4 Exercises

Exercise 17.1 ★

Nature (and sum, when telescoping) of:

\sum_{n\geq1} \frac{1}{n(n+1)}, \qquad \sum_{n\geq2} \ln\Bigl(1 - \frac{1}{n^2}\Bigr), \qquad \sum_{n\geq0} \frac{3^n + 4^n}{5^n} .

Solution

Solution of Exercise 17.1.

$\dfrac{1}{n(n+1)} = \dfrac1n - \dfrac{1}{n+1}$ : telescoping, $S_N = 1 - \frac{1}{N+1} \to 1$ . Convergent, sum $1$ .

$\ln\bigl(1 - \frac{1}{n^2}\bigr) = \ln\frac{(n-1)(n+1)}{n^2} = \ln\frac{n-1}{n} - \ln\frac{n}{n+1}$ : telescoping again, $S_N = \ln\frac12 - \ln\frac{N}{N+1} \to -\ln 2$ . Convergent, sum $-\ln 2$ .

$\dfrac{3^n + 4^n}{5^n} = \bigl(\frac35\bigr)^n + \bigl(\frac45\bigr)^n$ : two convergent geometric series, sum $\frac{1}{1 - 3/5} + \frac{1}{1 - 4/5} = \frac52 + 5 = \frac{15}{2}$ .

Exercise 17.2 ★

Nature of: $\;\sum \dfrac{n^2}{2^n}$ ; $\;\sum \dfrac{n!}{n^n}$ ; $\;\sum \dfrac{2^n\,n!}{n^n}$ ; $\;\sum \dfrac{3^n\,n!}{n^n}$ . (Ratio test; recall $\bigl(1 + \frac1n\bigr)^n \to \eu$ .)

Solution

Solution of Exercise 17.2.

Ratio test throughout.

$\frac{u_{n+1}}{u_n} = \frac{(n+1)^2}{2n^2} \to \frac12 < 1$ : convergent.

$\frac{u_{n+1}}{u_n} = \frac{(n+1)!\,n^n}{n!\,(n+1)^{n+1}} = \bigl(\frac{n}{n+1}\bigr)^n = \bigl(1 + \frac1n\bigr)^{-n} \to \frac1\eu < 1$ : convergent.

With the factor $2^n$ : ratio $\to \frac2\eu < 1$ : convergent.

With $3^n$ : ratio $\to \frac3\eu > 1$ : divergent (terms tend to $+\infty$ ).

Exercise 17.3 ★

Nature of: $\;\sum \sin\dfrac{1}{n^2}$ ; $\;\sum \Bigl(1 - \cos\dfrac1n\Bigr)$ ; $\;\sum \dfrac{1}{\sqrt{n(n+1)}}$ ; $\;\sum \dfrac{\ln n}{n^2}$ (compare with $n^{-3/2}$ ).

Solution

Solution of Exercise 17.3.

All nonnegative terms; use equivalents (Theorem 17.7).

$\sin\frac{1}{n^2} \sim \frac{1}{n^2}$ : convergent (Riemann $\alpha = 2$ ).

$1 - \cos\frac1n \sim \frac{1}{2n^2}$ : convergent.

$\frac{1}{\sqrt{n(n+1)}} \sim \frac1n$ : divergent.

$\frac{\ln n}{n^2} = \frac{1}{n^{3/2}}\cdot\frac{\ln n}{n^{1/2}}$ and $\frac{\ln n}{\sqrt n} \to 0$ (Proposition 4.6): so $\frac{\ln n}{n^2} \leq \frac{1}{n^{3/2}}$ for large $n$ : convergent.

Exercise 17.4 ★

Prove that $\sum_{n\geq1} \frac{1}{n^2}$ converges with sum $\leq 2$ , using $\frac{1}{n^2} \leq \frac{1}{n(n-1)}$ for $n \geq 2$ and a telescoping bound.

Solution

Solution of Exercise 17.4.

For $n \geq 2$ : $\frac{1}{n^2} \leq \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac1n$ . Hence

\sum_{n=1}^{N} \frac{1}{n^2} \leq 1 + \sum_{n=2}^{N} \Bigl(\frac{1}{n-1} - \frac1n\Bigr) = 1 + 1 - \frac1N < 2 :

partial sums increasing and bounded by $2$ : convergence (Theorem 17.7), sum $\leq 2$ . (The exact value $\frac{\pi^2}{6}$ is a second-year celebration.)

Exercise 17.5 ★★

Nature of $\;\sum \dfrac{(-1)^n}{\sqrt n}$ , of $\;\sum \dfrac{(-1)^n}{n + (-1)^n}$ (expand: the alternating test does not apply directly — why?), and of $\;\sum \sin\bigl(\pi\sqrt{n^2+1}\,\bigr)$ (reduce modulo $\pi$ : $\sqrt{n^2+1} = n + \frac{1}{2n} + O(n^{-3})$ ).

Solution

Solution of Exercise 17.5.

$\sum \frac{(-1)^n}{\sqrt n}$ : alternating with $\frac{1}{\sqrt n} \downarrow 0$ : convergent (Theorem 17.18); not absolutely ( $\alpha = \frac12 \leq 1$ ).

$\sum \frac{(-1)^n}{n + (-1)^n}$ : the sequence $\frac{1}{n + (-1)^n}$ is not decreasing ( $\frac{1}{n+1}$ then $\frac{1}{n}$ alternate badly), so the test does not apply directly. Expand:

\frac{(-1)^n}{n + (-1)^n} = \frac{(-1)^n}{n}\cdot\frac{1}{1 + \frac{(-1)^n}{n}} = \frac{(-1)^n}{n} - \frac{1}{n^2} + O\Bigl(\frac{1}{n^3}\Bigr):

the first series converges (alternating), $\sum \frac{1}{n^2}$ converges, the $O(n^{-3})$ converges absolutely: the sum of three convergent series converges.

$\sin\bigl(\pi\sqrt{n^2+1}\bigr)$ : write $\sqrt{n^2 + 1} = n + \frac{1}{2n} + \varepsilon_n$ with $\varepsilon_n = O(n^{-3})$ ; then, by $\pi$ -periodicity of $\sin$ up to sign,

\sin\bigl(\pi\sqrt{n^2+1}\bigr) = (-1)^n \sin\Bigl(\frac{\pi}{2n} + \pi\varepsilon_n\Bigr) .

Set $\theta_n = \frac{\pi}{2n} + \pi\varepsilon_n$ and $a_n = \sin\theta_n$ . For large $n$ , $\theta_n \in \intoo{0}{\frac\pi2}$ and

\theta_n - \theta_{n+1} = \frac{\pi}{2n(n+1)} + \pi(\varepsilon_n - \varepsilon_{n+1}) = \frac{\pi}{2n^2} + O\Bigl(\frac{1}{n^3}\Bigr) > 0

eventually, so $(\theta_n)$ decreases to $0$ ; since $\sin$ is increasing on $\intcc{0}{\frac\pi2}$ , $(a_n)$ decreases to $0$ as well. The alternating test applies: convergent — not absolutely, since $a_n \sim \frac{\pi}{2n}$ .

Exercise 17.6 ★★

For which $\alpha > 0$ does $\sum_{n \geq 2} \dfrac{1}{n (\ln n)^{\alpha}}$ converge? (Integral comparison; substitute $u = \ln t$ .)

Solution

Solution of Exercise 17.6.

$f(t) = \frac{1}{t(\ln t)^\alpha}$ is positive, continuous, decreasing on $\intco{2}{+\infty}$ . Substituting $u = \ln t$ :

\int_2^x \frac{\dd t}{t(\ln t)^\alpha} = \int_{\ln 2}^{\ln x} \frac{\dd u}{u^\alpha},

bounded as $x \to \infty$ iff $\alpha > 1$ (Theorem 17.10’s computation). By integral comparison: convergence iff $\alpha > 1$ . (These Bertrand-type series show how fine the boundary of convergence is: $n\ln n$ diverges, $n(\ln n)^{1.01}$ converges.)

Exercise 17.7 ★★

Let $u_n = \dfrac{1}{n} - \ln\Bigl(1 + \dfrac1n\Bigr)$ . Prove that $0 \leq u_n \leq \dfrac{1}{2n^2}$ , that $\sum u_n$ converges, and deduce the existence of Euler’s constant:

\gamma = \lim_{N \to \infty} \Bigl( \sum_{n=1}^{N} \frac 1n - \ln N \Bigr) .

Solution

Solution of Exercise 17.7.

By the tangent-line bounds of Exercise 14.3 rewritten via expansions: for $x = \frac1n \in \intoc{0}{1}$ , Taylor–Lagrange for $\ln(1+x)$ at order $1$ gives $\ln(1 + x) = x - \frac{x^2}{2(1 + c)^2}$ for some $c \in \intoo{0}{x}$ , so

0 \leq u_n = \frac1n - \ln\Bigl(1 + \frac1n\Bigr) \leq \frac{1}{2n^2} .

Comparison with the Riemann series: $\sum u_n$ converges. Its partial sum telescopes the logarithms:

\sum_{n=1}^{N} u_n = H_N - \ln(N+1)

(since $\sum_{n\leq N} \ln\frac{n+1}{n} = \ln(N+1)$ ). So $H_N - \ln(N+1)$ converges; adding $\ln\frac{N+1}{N} \to 0$ , the sequence $H_N - \ln N$ converges. Its limit is $\gamma \approx 0.5772$ .

Exercise 17.8 ★★

Compute the sums

\sum_{n=1}^{\infty} \frac{1}{n(n+2)} \qquad\text{and}\qquad \sum_{n=0}^{\infty} \frac{n}{2^n} .

(For the first: partial fractions. For the second: compute $\sum_{n=1}^{N} n x^{n-1}$ in closed form and let $N \to \infty$ at $x = \frac12$ .)

Solution

Solution of Exercise 17.8.

$\dfrac{1}{n(n+2)} = \dfrac{1/2}{n} - \dfrac{1/2}{n+2}$ : the partial sum telescopes with a lag of $2$ ,

S_N = \frac12\Bigl(1 + \frac12 - \frac{1}{N+1} - \frac{1}{N+2}\Bigr) \longrightarrow \frac34 .

$\sum \frac{n}{2^n}$ : for $\abs x < 1$ , differentiating the finite geometric sum and passing to the limit (all series here converge absolutely, ratio test): from $\sum_{n\geq0} x^n = \frac{1}{1-x}$ , one gets by direct computation with partial sums

\sum_{n=1}^{N} n x^{n-1} = \frac{1 - (N+1)x^N + N x^{N+1}}{(1 - x)^2} \xrightarrow[N\to\infty]{} \frac{1}{(1-x)^2} \quad (\abs x < 1),

(the boundary terms $N x^N \to 0$ ). At $x = \frac12$ : $\sum_{n\geq1} n\bigl(\frac12\bigr)^{n-1} = 4$ , so $\sum_{n\geq0} \frac{n}{2^n} = \frac12 \times 4 = 2$ .

Exercise 17.9 ★★★

(Cauchy condensation) Let $(u_n)$ be nonnegative and decreasing. Prove that

\sum_{n \geq 1} u_n \text{ converges} \iff \sum_{k \geq 0} 2^k\, u_{2^k} \text{ converges},

by comparing packets of terms between consecutive powers of $2$ . Recover from it the Riemann criterion and Exercise 17.6.

Solution

Solution of Exercise 17.9.

Group the terms of $\sum u_n$ in packets between powers of $2$ . Upper packets: for $2^k \leq n < 2^{k+1}$ there are $2^k$ terms, each $\leq u_{2^k}$ :

\sum_{n=1}^{2^{K+1}-1} u_n = \sum_{k=0}^{K} \sum_{n=2^k}^{2^{k+1}-1} u_n \leq \sum_{k=0}^{K} 2^k u_{2^k} .

Lower packets: each term of the same packet is $\geq u_{2^{k+1}}$ , so $\sum_{n=2^k}^{2^{k+1}-1} u_n \geq 2^k u_{2^{k+1}} = \frac12 \cdot 2^{k+1} u_{2^{k+1}}$ , whence

\sum_{n=1}^{2^{K+1}-1} u_n \geq \frac12 \sum_{k=1}^{K+1} 2^{k} u_{2^{k}} .

Both partial-sum comparisons go both ways (nonnegative terms, Theorem 17.7): the two series have the same nature.

Riemann: $u_n = n^{-\alpha}$ gives $2^k u_{2^k} = 2^{k(1-\alpha)}$ , a geometric series, convergent iff $2^{1 - \alpha} < 1$ iff $\alpha > 1$ . Bertrand (Exercise 17.6): $u_n = \frac{1}{n(\ln n)^\alpha}$ gives $2^k u_{2^k} = \frac{1}{(k\ln 2)^\alpha}$ , a Riemann series in $k$ : convergent iff $\alpha > 1$ .

Exercise 17.10 ★★★

Using the integral identity of Example 17.19 adapted to $\frac{1}{1+t^2}$ , prove Leibniz’s formula

\frac{\pi}{4} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = 1 - \frac13 + \frac15 - \frac17 + \cdots

with the error bound $\abs{R_N} \leq \frac{1}{2N+3}$ .

Solution

Solution of Exercise 17.10.

Finite geometric identity with ratio $-t^2$ :

\frac{1}{1 + t^2} = \sum_{k=0}^{n-1} (-1)^k t^{2k} + \frac{(-1)^n t^{2n}}{1 + t^2} .

Integrate over $\intcc{0}{1}$ (the left side integrates to $\arctan 1 = \frac\pi4$ , Proposition 4.10):

\frac{\pi}{4} = \sum_{k=0}^{n-1} \frac{(-1)^k}{2k+1} + (-1)^n \int_0^1 \frac{t^{2n}}{1+t^2}\,\dd t, \qquad 0 \leq \int_0^1 \frac{t^{2n}}{1+t^2}\,\dd t \leq \int_0^1 t^{2n}\dd t = \frac{1}{2n+1} .

Letting $n \to \infty$ proves the formula, and the displayed bound on the integral is exactly the remainder bound: after summing up to $N$ (i.e. $n = N + 1$ terms), $\abs{R_N} \leq \frac{1}{2N + 3}$ .

Exercise 17.11 ★★

Nature of $\displaystyle\sum_{n \geq 1} \frac{1}{n^{1 + 1/n}}$ . (Compute the limit of $n^{1/n}$ and find an equivalent of the general term: the Riemann test needs a fixed exponent.)

Solution

Solution of Exercise 17.11.

$n^{1/n} = \eu^{\frac{\ln n}{n}} \to \eu^0 = 1$ (Proposition 4.6). Hence

\frac{1}{n^{1 + 1/n}} = \frac{1}{n}\,\eu^{-\frac{\ln n}{n}} \sim \frac{1}{n} ,

and the equivalents test (Theorem 17.7) compares with the divergent harmonic series: divergent, although every exponent $1 + \frac1n$ exceeds $1$ . The Riemann criterion concerns a fixed exponent $\alpha$ ; an exponent sliding down to $1$ can lose all its margin, as here.

Exercise 17.12 ★★★

Let $(u_n)$ be nonnegative and decreasing with $\sum u_n$ convergent. Prove that $n\,u_n \to 0$ (bound $n\,u_{2n}$ by a slice $\sum_{k=n+1}^{2n} u_k$ and use the Cauchy criterion). Show that the converse fails, and that the monotonicity hypothesis cannot be removed.

Solution

Solution of Exercise 17.12.

Let $\varepsilon > 0$ . By the Cauchy criterion for the convergent series (Theorem 11.20 applied to the partial sums), there is $N$ with $\sum_{k=n+1}^{2n} u_k \leq \varepsilon$ for $n \geq N$ . By monotonicity each of these $n$ terms is $\geq u_{2n}$ :

n\,u_{2n} \leq \sum_{k=n+1}^{2n} u_k \leq \varepsilon \quad\Longrightarrow\quad 2n\,u_{2n} \leq 2\varepsilon ,

and for odd indices $(2n+1)\,u_{2n+1} \leq (2n+1)\,u_{2n} \leq 2\bigl(2n\,u_{2n}\bigr) \leq 4\varepsilon$ for $n \geq N$ : in both parities, $n u_n \to 0$ .

Converse false: $u_n = \frac{1}{n\ln n}$ has $n u_n = \frac{1}{\ln n} \to 0$ , yet the series diverges (Exercise 17.6, $\alpha = 1$ ). Monotonicity necessary: let $u_n = \frac1n$ when $n$ is a perfect square and $u_n = 2^{-n}$ otherwise: the series converges (the square terms sum like $\sum \frac{1}{k^2}$ , the rest geometrically), but $n u_n = 1$ along the squares.

17.5 Problem: Euler’s constant and the series that changes its sum

Problem 17.1

Weekend problem — $H_n = \ln n + \gamma + \frac{1}{2n} + O(n^{-2})$ , and rearranging $1 - \frac12 + \frac13 - \dots$ to $\frac{\ln 2}{2}$

Two stories share the harmonic series. First, the exact bookkeeping of its divergence: $H_n - \ln n$ converges to Euler’s constant $\gamma$ (Exercise 17.7), and this problem sharpens the statement into a two-sided law $\frac{1}{2(n+1)} \leq H_n - \ln n - \gamma \leq \frac{1}{2n}$ , certifying $\gamma = 0.5772\dots$ by hand. Second, the scandal of conditional convergence: the alternating harmonic series sums to $\ln 2$ (Example 17.19), yet the same terms, in a different order, sum to $\frac{\ln 2}{2}$ — or to $\ln 2 + \frac12\ln\frac pq$ for any $p, q$ , or to any real whatsoever (Riemann). The two stories are one: the rearranged sums are computed with the $\gamma$ -law.

Part I — $\gamma$ , bracketed. Set $a_n = H_n - \ln n$ and $b_n = H_n - \ln(n+1)$ .

Using $\frac{t}{1+t} \leq \ln(1 + t) \leq t$ , show that $(a_n)$ decreases, $(b_n)$ increases, and that they are adjacent; their common limit is $\gamma$ , with $b_n \leq \gamma \leq a_n$ for every $n$ .
Numerical first shot: from $H_{10} = 2.928968\dots$ , bracket $\gamma$ between $b_{10} = 0.5311$ and $a_{10} = 0.6264$ . How large an $n$ would this crude bracket need for four decimals?
Show the exact tail representation $a_n - \gamma = \sum_{k \geq n} w_k$ (limit of partial sums), where

$w_k = a_k - a_{k+1} = \ln\Bigl(1 + \frac1k\Bigr) - \frac{1}{k+1} = \int_k^{k+1} \frac{(k + 1 - t)}{t\,(k+1)}\,\dd t ,$

and deduce from the integral form the two-sided bound $\dfrac{1}{2(k+1)^2} \leq w_k \leq \dfrac{1}{2k(k+1)}$ .

Part II — The $\frac{1}{2n}$ law.

Sum the bounds of question 3 (both sides telescope or compare to telescopes) and conclude the law:

$\frac{1}{2(n+1)} \;\leq\; H_n - \ln n - \gamma \;\leq\; \frac{1}{2n} \qquad (n \geq 1).$
Deduce $H_n = \ln n + \gamma + \frac{1}{2n} + O\bigl(\frac{1}{n^2}\bigr)$ ; precisely, show that $\gamma_n = H_n - \ln n - \frac{1}{2n}$ satisfies $-\frac{1}{2n(n+1)} \leq \gamma_n - \gamma \leq 0$ .
Certify four decimals with $n = 100$ : given $H_{100} = 5.1873775\dots$ , compute $\gamma_{100} = 0.577207\dots$ and conclude $\gamma = 0.5772 \pm 5\cdot10^{-5}$ (true value $0.5772156\dots$ ).
Two dividends of the law, both needed later: as $m \to \infty$ ,

$H_{2m} - H_m = \ln 2 - \frac{1}{4m} + O\Bigl(\frac{1}{m^2}\Bigr), \qquad \sum_{j=1}^{m} \frac{1}{2j-1} = \frac{\ln m}{2} + \ln 2 + \frac\gamma2 + o(1) ,$

the second via $\sum_{j \leq m} \frac{1}{2j-1} = H_{2m} - \frac12 H_m$ , and likewise $\sum_{j=1}^{m} \frac{1}{2j} = \frac{\ln m}{2} + \frac\gamma2 + o(1)$ .

Part III — The alternating harmonic series, to second order.

Show (induction, or grouping) the identity $\sum_{k=1}^{2m} \frac{(-1)^{k-1}}{k} = H_{2m} - H_m$ , and deduce both the sum $\ln 2$ (again) and the exact speed:

$\sum_{k=1}^{2m} \frac{(-1)^{k-1}}{k} = \ln 2 - \frac{1}{4m} + O\Bigl(\frac{1}{m^2}\Bigr) .$
Deduce the asymptotic error of the alternating harmonic series at any index: $S - S_N \sim \frac{(-1)^N}{2N}$ — twice smaller than the worst-case bound $a_{N+1} \approx \frac1N$ of Theorem 17.18.
(Acceleration for free) Show that the averaged sums $\tilde S_N = \frac{S_N + S_{N+1}}{2}$ satisfy $\tilde S_N = \ln 2 + O\bigl(\frac{1}{N^2}\bigr)$ . Check: $S_{10} = 0.64563$ , $S_{11} = 0.73654$ , $\tilde S_{10} = 0.69109$ , against $\ln 2 = 0.69315$ : one average buys two decimal places.
Explain in two sentences why no such trick can help a positive divergent-tail phenomenon like question 2’s bracket: the alternating error oscillates (sign $(-1)^N$ ), so averaging cancels its leading term, while the $\gamma$ -bracket error $\frac{1}{2n}$ has constant sign. (Averaging $a_n$ and $b_n$ does help: relate $\frac{a_n + b_n}{2}$ to the midpoint estimate $H_n - \ln\bigl(n + \frac12\bigr)$ and show its error is $O\bigl(\frac{1}{n^2}\bigr)$ .)

Part IV — Rigidity and its failure.

Show that the positive part $\sum \frac{1}{2j-1}$ and the negative part $\sum \frac{1}{2j}$ of the alternating harmonic series both diverge — the signature of conditional convergence.
Prove the general statement behind question 12: if $\sum u_n$ converges but $\sum \abs{u_n}$ diverges, then the series of positive parts $\sum u_n^+$ and of negative parts $\sum u_n^-$ both diverge (from $u_n^\pm = \frac{\abs{u_n} \pm u_n}{2}$ : if one converged, so would the other, hence $\sum\abs{u_n}$ ). This inexhaustible reservoir of positive and negative mass is what Riemann’s recipe will spend.
(Rigidity) Prove: if $\sum u_n$ converges absolutely and $\sigma \colon \N \to \N$ is a bijection, then $\sum u_{\sigma(n)}$ converges to the same sum (for $N$ large the first $M$ rearranged terms contain $u_0, \dots, u_N$ ; compare partial sums through the tail $\sum_{n > N}\abs{u_n}$ ).
(Riemann’s recipe) Let $t \in \R$ . Describe the greedy rearrangement of the alternating harmonic series: take positive terms $1, \frac13, \frac15, \dots$ until the partial sum first exceeds $t$ , then negative terms $-\frac12, -\frac14, \dots$ until it first drops below $t$ , and repeat. Show that every term is used exactly once, that after the first crossing the partial sums stay within the last used term of $t$ , and conclude that the rearranged series converges to $t$ : any prescribed sum is attainable.

Part V — The $(p, q)$ formula. Fix integers $p, q \geq 1$ . Rearrange the alternating harmonic series in blocks: $p$ positive terms (the next odd reciprocals), then $q$ negative terms (the next even reciprocals), and repeat.

Check that this is a genuine rearrangement (every term exactly once), and that for $(p, q) = (1, 2)$ it reads

$1 - \frac12 - \frac14 + \frac13 - \frac16 - \frac18 + \frac15 - \frac1{10} - \frac1{12} + \dots$
(The exact halving) For $(p, q) = (1, 2)$ , prove the block identity

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

and deduce the exact relation $T_{3K} = \frac12 S_{2K}$ between the rearranged partial sums and the original ones: the halving of the sum is visible at every finite stage, not only in the limit.
Show that the partial sum after $K$ complete blocks equals $\sum_{j=1}^{pK} \frac{1}{2j-1} - \sum_{j=1}^{qK} \frac{1}{2j}$ , and compute its limit with question 7:

$\ln 2 + \frac12 \ln\frac pq .$
Control the partial sums inside a block (the terms tend to $0$ ) and conclude that the $(p, q)$ -rearranged series converges to $\ln 2 + \frac12\ln \frac pq$ . In particular $(1, 2)$ gives $\frac{\ln 2}{2}$ : verify against the first nine terms, $T_9 = 0.3083$ , creeping toward $0.3466$ .
Sanity checks and range: $(1,1)$ recovers $\ln 2$ ; $(2,1)$ gives $\frac32\ln 2$ ; which sums are reachable by $(p, q)$ -blocks, and how does this countable menu compare with Riemann’s full carte (question 14)?

Part VI — Epilogue: $\gamma$ at work, and synthesis.

Identify the sum of the convergent series $\sum_{k\geq1} \bigl(\frac1k - \ln\frac{k+1}{k}\bigr)$ (Exercise 17.7): show it equals $\gamma$ .
Run Riemann’s recipe (question 14) for the target $t = 1$ and list the first twelve terms produced ( $1, \frac13, -\frac12, \frac15, -\frac14, \frac17, \frac19, -\frac16, \frac1{11}, \frac1{13}, -\frac18, \frac1{15}$ ), computing the partial sum ( $\approx 0.980$ ) — watch the algorithm breathe around its target.
Show that some rearrangement of the alternating harmonic series diverges to $+\infty$ (blocks of positive terms long enough to gain $1$ each time, using question 12, separated by single negative terms).
Sharpen Example 17.11 with the $\gamma$ -law: show that the first index with $H_N \geq 20$ satisfies $N = \eu^{\,20 - \gamma}\,(1 + o(1)) \approx 2.7\cdot10^{8}$ — Euler’s constant is exactly the correction the crude framing was missing.
Synthesis, one sentence each: (i) the $\gamma$ -law and what each of its three pieces ( $\ln n$ , $\gamma$ , $\frac{1}{2n}$ ) contributes; (ii) why conditional convergence makes the sum order-dependent while absolute convergence forbids it; (iii) how the $(p,q)$ formula was a computation with the $\gamma$ -law rather than an abstract existence claim; (iv) where these threads continue — power series and products of series in the Year 2 volume, and the Year 3 volume’s weekend problem on Stirling’s formula, where the same sum-versus-integral bookkeeping runs at full power.

Solution

Solution of Problem 17.1.

1. $a_{n+1} - a_n = \frac{1}{n+1} - \ln\frac{n+1}{n} \leq 0$ because $\ln(1 + \frac1n) \geq \frac{1/n}{1 + 1/n} = \frac{1}{n+1}$ ; and $b_{n+1} - b_n = \frac{1}{n+1} - \ln\frac{n+2}{n+1} \geq 0$ because $\ln(1 + \frac{1}{n+1}) \leq \frac{1}{n+1}$ . Their gap $a_n - b_n = \ln(1 + \frac1n) \to 0$ : adjacent (Theorem 11.11), with common limit $\lim a_n = \gamma$ (Exercise 17.7), and $b_n \leq \gamma \leq a_n$ .

2. $b_{10} = 2.928968 - \ln 11 = 0.5311$ and $a_{10} = 2.928968 - \ln 10 = 0.6264$ : so $\gamma \in \intcc{0.5311}{0.6264}$ . The gap is $\ln 1.1 \approx 0.095$ and shrinks like $\frac1n$ : four decimals ( $\text{gap} \leq 10^{-4}$ ) would need $n \approx 10^4$ — the brackets are correct but slow.

3. Telescoping $a_n - a_{m+1} = \sum_{k=n}^{m} w_k$ and letting $m \to \infty$ : $a_n - \gamma = \sum_{k\geq n} w_k$ (limit of partial sums). Moreover

w_k = \int_k^{k+1} \frac{\dd t}{t} - \frac{1}{k+1} = \int_k^{k+1} \Bigl(\frac1t - \frac{1}{k+1}\Bigr)\dd t = \int_k^{k+1} \frac{k + 1 - t}{t\,(k+1)}\,\dd t .

On $\intcc{k}{k+1}$ : $\frac{1}{(k+1)^2} \leq \frac{1}{t(k+1)} \leq \frac{1}{k(k+1)}$ , and $\int_k^{k+1}(k + 1 - t)\dd t = \frac12$ : hence $\frac{1}{2(k+1)^2} \leq w_k \leq \frac{1}{2k(k+1)}$ .

4. Upper: $\sum_{k \geq n} \frac{1}{2k(k+1)} = \frac12\sum_{k\geq n}\bigl(\frac1k - \frac{1}{k+1}\bigr) = \frac{1}{2n}$ (telescoping). Lower: $\frac{1}{2(k+1)^2} \geq \frac{1}{2(k+1)(k+2)}$ , whose sum telescopes to $\frac{1}{2(n+1)}$ . With question 3:

\frac{1}{2(n+1)} \leq H_n - \ln n - \gamma \leq \frac{1}{2n} .

5. Subtract $\frac{1}{2n}$ : $\gamma_n - \gamma \in \intcc{\frac{1}{2(n+1)} - \frac{1}{2n}}{0} = \intcc{-\frac{1}{2n(n+1)}}{0}$ : the corrected estimate is exact to $O\bigl(\frac{1}{n^2}\bigr)$ , and always from below.

6. $\gamma_{100} = 5.1873775 - \ln 100 - 0.005 = 0.5772073$ , with $0 \leq \gamma - \gamma_{100} \leq \frac{1}{20200} < 5\cdot10^{-5}$ : hence $0.577207 \leq \gamma \leq 0.577257$ , i.e. $\gamma = 0.5772 \pm 5\cdot10^{-5}$ (true value $0.5772156\dots$ ) — four certified decimals from a hundred terms, against ten thousand for question 2.

7. First dividend:

H_{2m} - H_m = \Bigl(\ln 2m + \gamma + \frac{1}{4m}\Bigr) - \Bigl(\ln m + \gamma + \frac{1}{2m}\Bigr) + O\Bigl(\frac{1}{m^2}\Bigr) = \ln 2 - \frac{1}{4m} + O\Bigl(\frac{1}{m^2}\Bigr).

Second: the even reciprocals up to $2m$ sum to $\frac12 H_m$ , so $\sum_{j=1}^{m}\frac{1}{2j-1} = H_{2m} - \frac12 H_m = \frac12 \ln m + \ln 2 + \frac\gamma2 + o(1)$ , and $\sum_{j=1}^m \frac{1}{2j} = \frac12\ln m + \frac\gamma2 + o(1)$ .

8. Splitting off the even terms twice: $\sum_{k=1}^{2m}\frac{(-1)^{k-1}}{k} = H_{2m} - 2\cdot\frac12 H_m = H_{2m} - H_m$ . By question 7 this equals $\ln 2 - \frac{1}{4m} + O(m^{-2})$ : the sum is $\ln 2$ (Example 17.19 again) with its speed.

9. For $N = 2m$ : $S - S_N = \frac{1}{4m} + O(m^{-2}) = \frac{1}{2N} + O(N^{-2})$ . For $N = 2m + 1$ : $S_{N} = S_{2m} + \frac{1}{2m+1}$ , so

S - S_N = \Bigl(\frac{1}{4m} - \frac{1}{2m+1}\Bigr) + O\Bigl(\frac{1}{m^2}\Bigr) = -\frac{1}{4m} + O\Bigl(\frac{1}{m^2}\Bigr) = -\frac{1}{2N} + O\Bigl(\frac{1}{N^2}\Bigr).

In both cases $S - S_N \sim \frac{(-1)^N}{2N}$ : half the worst-case bound $a_{N+1}$ , with a known, alternating sign.

10. Averaging kills the oscillating leading term:

S - \tilde S_N = \frac{(S - S_N) + (S - S_{N+1})}{2} = \frac{(-1)^N}{2}\Bigl(\frac{1}{2N} - \frac{1}{2(N+1)}\Bigr) + O\Bigl(\frac{1}{N^2}\Bigr) = O\Bigl(\frac{1}{N^2}\Bigr).

Numerically: $S_{10} = 0.645635$ , $S_{11} = 0.736544$ , $\tilde S_{10} = 0.691089$ , and $\ln 2 = 0.693147$ : the error drops from $4.8\cdot10^{-2}$ to $2.1\cdot10^{-3}$ — one addition, twenty times better.

11. The alternating error changes sign at every step, so consecutive partial sums straddle the limit and their mean cancels the first-order term; the bracket error $H_n - \ln n - \gamma \approx \frac{1}{2n}$ has constant sign, so no averaging along $n$ can cancel it. Averaging the two brackets does help: $\frac{a_n + b_n}{2} = H_n - \ln\sqrt{n(n+1)}$ , and since $\ln\sqrt{n(n+1)} = \ln\bigl(n + \frac12\bigr) + O(n^{-2})$ ,

H_n - \ln\Bigl(n + \frac12\Bigr) = \Bigl(H_n - \ln n - \frac{1}{2n}\Bigr) + \frac{1}{8n^2} + O\Bigl(\frac{1}{n^3}\Bigr) = \gamma + O\Bigl(\frac{1}{n^2}\Bigr)

(question 5 and $\ln(1 + \frac{1}{2n}) = \frac{1}{2n} - \frac{1}{8n^2} + O(n^{-3})$ ). Check at $n = 10$ : $H_{10} - \ln 10.5 = 0.57759$ , already within $4\cdot10^{-4}$ of $\gamma$ .

12. $\sum_{j\leq m} \frac{1}{2j-1} \geq \sum_{j \leq m} \frac{1}{2j} = \frac12 H_m \to +\infty$ : both the positive and the negative part of the alternating harmonic series diverge.

13. Write $u_n^\pm = \frac{\abs{u_n} \pm u_n}{2} \geq 0$ , so $u_n = u_n^+ - u_n^-$ and $\abs{u_n} = u_n^+ + u_n^-$ . If $\sum u_n^+$ converged, then $\sum u_n^- = \sum (u_n^+ - u_n)$ would converge (difference of convergent series), hence $\sum \abs{u_n}$ too: contradiction with conditional convergence. By symmetry both $\sum u_n^\pm$ diverge (to $+\infty$ ): an infinite reservoir of positive and of negative mass.

14. Let $S = \sum u_n$ , $\varepsilon > 0$ , and $N$ with $\sum_{n > N} \abs{u_n} \leq \varepsilon$ (Cauchy criterion for $\sum\abs{u_n}$ ). Let $M_0$ be large enough that $\sigma(\{0, \dots, M_0\}) \supseteq \{0, \dots, N\}$ . For $M \geq M_0$ , the difference $\sum_{m \leq M} u_{\sigma(m)} - \sum_{n \leq N} u_n$ is a finite sum of distinct terms $u_n$ with $n > N$ , hence of absolute value $\leq \varepsilon$ ; and $\abs{S - \sum_{n\leq N} u_n} \leq \varepsilon$ as well. So the rearranged partial sums are within $2\varepsilon$ of $S$ eventually: $\sum u_{\sigma(n)} = S$ . Absolute convergence is rearrangement-proof.

15. Each phase of the greedy procedure ends after finitely many terms, because the remaining positive (respectively negative) terms alone have divergent partial sums (question 12): the running sum must eventually cross $t$ . The procedure therefore alternates infinitely many finite phases, consuming the positive terms in order and the negative terms in order: every term is used exactly once — a rearrangement. After the first crossing, between two consecutive crossings the partial sums move monotonically toward $t$ , and at a crossing they overshoot by at most the term just added; since the terms used at the $j$ -th crossing have index at least $j$ in their class, these overshoots tend to $0$ . Hence the partial sums converge to $t$ : every real number is the sum of some rearrangement.

16. The positive slots receive $\frac{1}{2j-1}$ for $j = 1, 2, \dots$ in order, the negative slots $\frac{1}{2j}$ in order: every term of the alternating harmonic series appears exactly once. For $(p, q) = (1, 2)$ , the blocks are $\bigl(1, -\frac12, -\frac14\bigr)$ , $\bigl(\frac13, -\frac16, -\frac18\bigr)$ , $\bigl(\frac15, -\frac1{10}, -\frac1{12}\bigr)$ , … — the displayed series.

17. Since $\frac{1}{4k-2} = \frac12\cdot\frac{1}{2k-1}$ :

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

Summing over $k = 1, \dots, K$ : $T_{3K} = \frac12 \sum_{k=1}^{K}\bigl(\frac{1}{2k-1} - \frac{1}{2k}\bigr) = \frac12 S_{2K}$ : at every third partial sum, the rearranged series is exactly half the original one.

18. After $K$ complete blocks, the rearranged partial sum is $\sum_{j=1}^{pK}\frac{1}{2j-1} - \sum_{j=1}^{qK}\frac{1}{2j}$ , and question 7 evaluates it:

\Bigl(\frac{\ln(pK)}{2} + \ln 2 + \frac\gamma2\Bigr) - \Bigl(\frac{\ln(qK)}{2} + \frac\gamma2\Bigr) + o(1) = \ln 2 + \frac12\ln\frac pq + o(1) :

the $\gamma$ s cancel, the $\ln K$ s cancel, the ratio $\frac pq$ survives.

19. A partial sum inside block $K + 1$ differs from the $K$ -block sum by at most $p + q$ terms, each of absolute value $\leq \frac{1}{2qK}$ -ish, hence by $O\bigl(\frac1K\bigr) \to 0$ : the full sequence of partial sums has the same limit $\ln 2 + \frac12\ln\frac pq$ . For $(1, 2)$ : $\ln 2 + \frac12\ln\frac12 = \frac{\ln 2}{2} = 0.34657\dots$ , and indeed $T_9 = 0.30833$ creeps toward it: by question 17, $T_{3K} = \frac12 S_{2K}$ converges with exactly half the alternating-harmonic error. Same terms, half the sum.

20. $(1,1)$ : $\ln 2 + \frac12\ln 1 = \ln 2$ — the original order, consistency. $(2,1)$ : $\frac32\ln 2 \approx 1.0397$ . The $(p,q)$ menu reaches exactly the countable dense family $\ln 2 + \frac12\ln r$ , $r \in \Q_{>0}$ ; Riemann’s greedy recipe (question 15) reaches every real. Structure buys formulas; greed buys totality.

21. The partial sums telescope: $\sum_{k=1}^{N} \bigl(\frac1k - \ln\frac{k+1}{k}\bigr) = H_N - \ln(N+1) = b_N \to \gamma$ : the series of Exercise 17.7 sums exactly to Euler’s constant.

22. Greedy for $t = 1$ : the first positive term brings the sum exactly to $1$ , not beyond it, so a second positive is taken to cross: $1, \frac13$ (sum $1.3333 > 1$ ), then $-\frac12$ ( $0.8333$ ), $\frac15$ ( $1.0333$ ), $-\frac14$ ( $0.7833$ ), $\frac17, \frac19$ ( $1.0373$ ), $-\frac16$ ( $0.8706$ ), $\frac1{11}, \frac1{13}$ ( $1.0384$ ), $-\frac18$ ( $0.9134$ ), $\frac1{15}$ ( $0.9801$ ), … — the sums breathe around $1$ with ever smaller amplitude, two positives now needed per cycle since the negatives are larger.

23. Build blocks: at stage $j$ , append enough unused positive terms to raise the partial sum by at least $1$ (possible: the remaining positive terms have divergent sums, question 12), then append the single negative term $-\frac{1} {2j}$ . Every positive term is eventually used (each stage uses at least one), every negative one too (one per stage): a rearrangement. Each stage changes the sum by $\geq 1 - \frac{1}{2j} \geq \frac12$ : the partial sums exceed $\frac{j}{2}$ after stage $j$ and the increments within a stage are positive except the last, bounded by $\frac{1}{2j} \to 0$ : divergence to $+\infty$ .

24. By the law, $H_N \geq 20 \iff \ln N \geq 20 - \gamma - \frac{1}{2N} + O(N^{-2})$ : the threshold $N^*$ satisfies $\ln N^* = 20 - \gamma + o(1)$ , i.e. $N^* = \eu^{20 - \gamma}(1 + o(1)) \approx \eu^{19.4228} \approx 2.72\cdot10^{8}$ — inside the crude window $\intcc{1.8\cdot10^8}{4.9\cdot10^8}$ of Example 17.11, and pinned by $\gamma$ .

25. (i) In $H_n = \ln n + \gamma + \frac{1}{2n} + O(n^{-2})$ : the $\ln n$ is the integral, $\gamma$ the price of replacing a sum by an integral (a genuinely new constant of analysis), and $\frac{1}{2n}$ the first correction — the trapezoid’s shadow. (ii) Conditional convergence leans on cancellation between two infinite reservoirs (question 13), so reordering re-weights the reservoirs; absolute convergence has finite total mass, and question 14’s tail estimate is order-blind. (iii) The $(p,q)$ sums were computed: the $\gamma$ -law turned each rearranged partial sum into $\ln 2 + \frac12\ln\frac pq + o(1)$ , with $\gamma$ itself canceling — an asymptotic bookkeeping exercise, not an abstract argument. (iv) Next: products and unconditional summability for power series in the Year 2 volume; and the Year 3 volume’s weekend problem on Stirling’s formula, where sum-versus-integral bookkeeping, pushed one order further, produces $\sqrt{2\pi}$ itself.