---
title: "Numerical Series"
book: "University Mathematics — Year 1"
subject: math
language: en
chapter: 17
exercises: 12
source: https://one-course.com/books/math/3/en/chapter/17-numerical-series
---

# Chapter 17 — Numerical Series

Summing infinitely many numbers means taking the limit of the partial sums — nothing more, nothing less. This chapter [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) up the definitions and the convergence tests usable in first year: comparison and equivalents for positive terms, the [ratio test](#thm-b1-series-ratio), the [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) comparison giving the [Riemann series](#thm-b1-series-riemann), [absolute convergence](#thm-b1-series-absolute), and the [alternating series](#thm-b1-series-alternating) theorem. The finer theory (products of [series](#def-b1-series-def), summation by packets, [series](#def-b1-series-def) 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](#def-b1-series-def) converges iff $\abs q < 1$ ([Exercise 11.3](https://one-course.com/books/math/3/en/chapter/11-sequences#exo-b1-seq-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](#def-b1-series-def):

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

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](https://one-course.com/books/math/3/en/chapter/10-real-numbers#pb-b1-reals-1), which this chapter’s language finally states in one line: a [decimal expansion](https://one-course.com/books/math/3/en/chapter/10-real-numbers#pb-b1-reals-1) is a convergent [series](#def-b1-series-def), eventually periodic exactly when its sum is rational. The digit machinery of Chapter 10, built there with bare suprema, was [series](#def-b1-series-def) theory traveling incognito.

**Proposition 17.4 (First facts).**

1. If $\sum u_n$ converges, then $u_n \to 0$ . (The converse is *false* : the harmonic [series](#def-b1-series-def) .)
2. Linearity: convergent [series](#def-b1-series-def) add and scale, with the expected sums.
3. (Telescoping) $\sum (v_{n+1} -  v_n)$ converges iff $(v_n)$ converges, with sum $\lim v_n -  v_0$ .
4. 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](#def-b1-series-def) has $u_n = \frac1n \to 0$ yet diverges ([Exercise 11.5](https://one-course.com/books/math/3/en/chapter/11-sequences#exo-b1-seq-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](#ex-b1-series-geometric) 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](https://one-course.com/books/math/3/en/chapter/4-standard-functions#def-b1-functions-arc) in [Problem 16.1](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#pb-b1-taylor-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](https://one-course.com/books/math/3/en/chapter/9-rational-fractions#ch-b1-fractions)):

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

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](#def-b1-series-def) have the same nature.

**Proof.** Monotone limit theorem ([Theorem 11.9](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-monotone)) 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](#def-b1-series-def): 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](#def-b1-series-def) of [Exercise 17.3](#exo-b1-series-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](#def-b1-series-def) converges. The whole decision took one equivalent and one lookup — provided the terms are nonnegative, which they are. The closing insight: for positive [series](#def-b1-series-def), 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](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor)), never in the summation.

**Theorem 17.10 (Integral comparison; Riemann series).**

Let $f$ be [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-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](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) $\int_1^x f$ bound each other within the constant $f(1)$, and both are nondecreasing, so one is bounded iff the other is ([Theorem 17.7](#thm-b1-series-positive)). 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](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) 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](#def-b1-series-def) 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](#pb-b1-series-1).) The closing insight: the [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) 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](#ex-b1-series-geometric). 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](#prop-b1-series-first) (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](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#thm-b1-taylor-lagrange)) 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](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#exo-b1-integration-9) (1) used the same fact). The same argument sums the $\sin$, $\cos$, $\sinh$, $\cosh$ [series](#def-b1-series-def) 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](#thm-b1-series-ratio) 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](https://one-course.com/books/math/3/en/chapter/2-counting#def-b1-counting-objects), 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](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#pb-b1-integration-1). The closing insight: quotients of factorials are exactly what the [ratio test](#thm-b1-series-ratio) 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](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-complete)):

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

which is small for large $N$ since the partial sums of $\sum\abs{u_n}$ form a [Cauchy sequence](https://one-course.com/books/math/3/en/chapter/11-sequences#def-b1-seq-cauchy). 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](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) in $\intcc{-1}{1}$, [Exercise 11.12](https://one-course.com/books/math/3/en/chapter/11-sequences#exo-b1-seq-12)), and no alternating structure is in sight. [Absolute convergence](#thm-b1-series-absolute) rescues everything at once: $\bigl|\frac{\sin n}{n^2}\bigr| \leq \frac{1}{n^2}$, a convergent scale, so the [series](#def-b1-series-def) 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](#thm-b1-series-absolute) is the only tool of this chapter that never asks how the signs are organized; try it first ([Method 17.21](#met-b1-series-decide)), and reserve the delicate tests for the [series](#def-b1-series-def) that fail it.

**Theorem 17.18 (Alternating series test).**

Let $(a_n)$ be decreasing with $a_n \to 0$. Then the [alternating series](#thm-b1-series-alternating) $\sum (-1)^n a_n$ 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](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-adjacent) they share a limit $S$, which the two [subsequences](https://one-course.com/books/math/3/en/chapter/11-sequences#def-b1-seq-subsequence) criterion ([Proposition 11.14](https://one-course.com/books/math/3/en/chapter/11-sequences#prop-b1-seq-subsequences)) 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](#def-b1-series-def)). 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](#thm-b1-series-alternating) 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.](https://one-course.com/images/onecourse/chapters/math-3/b1-series/fig-4fb812dd779c.svg)

*The partial sums $S_N$ of the alternating harmonic [series](#def-b1-series-def) $1 - \frac12 + \frac13 - \cdots$ 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](#thm-b1-series-alternating) made visible — and the slow closing of the pincers ($\abs{S_N - \ln 2} \approx
\frac{1}{2N}$, weekend problem [Problem 17.1](#pb-b1-series-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](#def-b1-series-def) 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](#def-b1-series-def) is the eternal counterexample; the converse direction ([Proposition 17.4](#prop-b1-series-first) (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](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) comparison. (iv) *Alternating needs decreasing*: $\sum
\frac{(-1)^n}{n + (-1)^n}$ looks alternating and is handled only by expansion ([Exercise 17.5](#exo-b1-series-5)); the weekend problem of [Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor) (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](#def-b1-series-def) 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](#pb-b1-series-1)).

**Method 17.21 (Deciding the nature of a series).**

1. Does $u_n \to 0$ ? If not, divergence, stop.
2. Nonnegative terms: seek an equivalent of $u_n$ (expansions, [Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor) !), compare with Riemann or geometric scales; factorials and powers call for the [ratio test](#thm-b1-series-ratio) ; decreasing $f(n)$ calls for [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) comparison.
3. Signs vary: try [absolute convergence](#thm-b1-series-absolute) 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](#exo-b1-series-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](#prop-b1-series-first) (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](https://one-course.com/books/math/3/en/chapter/11-sequences#ch-b1-seq)), itself the completeness axiom of [Chapter 10](https://one-course.com/books/math/3/en/chapter/10-real-numbers#ch-b1-reals); [absolute convergence](#thm-b1-series-absolute) is the Cauchy criterion; the [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) test is [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration)’s framing of areas; equivalents of general terms are [Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor)’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](#pb-b1-series-1)) 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](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) 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](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-open) three doors. In the Year 2 volume, [series](#def-b1-series-def) acquire a variable ($\sum
a_n x^n$: power [series](#def-b1-series-def), with their radius of convergence) and then a function-valued theory (Fourier [series](#def-b1-series-def)); 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](#def-b1-series-def), and [absolute convergence](#thm-b1-series-absolute) is what makes them well defined. And the [Riemann series](#thm-b1-series-riemann) $\sum n^{-s}$, pushed to complex $s$, becomes the zeta function — the single most studied [series](#def-b1-series-def) 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 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](#ex-b1-series-geometric), 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](#thm-b1-series-ratio); recall $\bigl(1 + \frac1n\bigr)^n \to \eu$.)*

**Solution of Exercise 17.2.**

[Ratio test](#thm-b1-series-ratio) 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 of Exercise 17.3.**

All nonnegative terms; use equivalents ([Theorem 17.7](#thm-b1-series-positive)).

$\sin\frac{1}{n^2} \sim \frac{1}{n^2}$: 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](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-powerrules)): 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 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](#thm-b1-series-positive)), 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 of Exercise 17.5.**

$\sum \frac{(-1)^n}{\sqrt n}$: alternating with $\frac{1}{\sqrt n}
\downarrow 0$: convergent ([Theorem 17.18](#thm-b1-series-alternating)); 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](#def-b1-series-def) converges (alternating), $\sum \frac{1}{n^2}$ converges, the $O(n^{-3})$ converges absolutely: the sum of three convergent [series](#def-b1-series-def) 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](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) comparison; substitute $u = \ln t$.)*

**Solution of Exercise 17.6.**

$f(t) = \frac{1}{t(\ln t)^\alpha}$ is positive, [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-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](#thm-b1-series-riemann)’s computation). By [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) comparison: convergence iff $\alpha > 1$. (These *Bertrand-type* [series](#def-b1-series-def) show how fine the [boundary](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure) 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](#pb-b1-series-1):

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

**Solution of Exercise 17.7.**

By the tangent-line bounds of [Exercise 14.3](https://one-course.com/books/math/3/en/chapter/14-differentiation#exo-b1-derivative-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](#thm-b1-series-riemann): $\sum u_n$ 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 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](#def-b1-series-def) here converge absolutely, [ratio test](#thm-b1-series-ratio)): 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](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure) 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](#exo-b1-series-6).

**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](#thm-b1-series-positive)): the two [series](#def-b1-series-def) have the same nature.

Riemann: $u_n = n^{-\alpha}$ gives $2^k u_{2^k} =
2^{k(1-\alpha)}$, a [geometric series](#ex-b1-series-geometric), convergent iff $2^{1 - \alpha}
< 1$ iff $\alpha > 1$. Bertrand ([Exercise 17.6](#exo-b1-series-6)): $u_n = \frac{1}{n(\ln n)^\alpha}$ gives $2^k u_{2^k} = \frac{1}{(k\ln 2)^\alpha}$, a [Riemann series](#thm-b1-series-riemann) in $k$: convergent iff $\alpha > 1$.

**Exercise 17.10 ★★★.**

Using the [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) identity of [Example 17.19](#ex-b1-series-ln2) 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 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](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-arcderiv)):

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

Letting $n \to \infty$ proves the formula, and the displayed bound on the [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) 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 of Exercise 17.11.**

$n^{1/n} = \eu^{\frac{\ln n}{n}} \to \eu^0 = 1$ ([Proposition 4.6](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-powerrules)). 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](#thm-b1-series-positive)) compares with the divergent harmonic [series](#def-b1-series-def): *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 of Exercise 17.12.**

Let $\varepsilon > 0$. By the Cauchy criterion for the convergent [series](#def-b1-series-def) ([Theorem 11.20](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-complete) 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](#def-b1-series-def) diverges ([Exercise 17.6](#exo-b1-series-6), $\alpha = 1$). Monotonicity necessary: let $u_n = \frac1n$ when $n$ is a perfect square and $u_n =
2^{-n}$ otherwise: the [series](#def-b1-series-def) 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](#def-b1-series-def). First, the exact bookkeeping of its divergence: $H_n - \ln n$ converges to [Euler’s constant](#pb-b1-series-1) $\gamma$ ([Exercise 17.7](#exo-b1-series-7)), and this problem sharpens the [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-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](#def-b1-series-def) sums to $\ln 2$ ([Example 17.19](#ex-b1-series-ln2)), 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 $b_n = H_n - \ln(n+1)$.

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 $n$ .
2. 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?
3. 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](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) 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.**

4. Sum the bounds of question 3 (both sides telescope or compare to telescopes) and conclude the law: $$\frac{1}{2(n+1)} \;\leq\; H_n - \ln n - \gamma \;\leq\;  \frac{1}{2n} \qquad (n \geq 1).$$
5. 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 $-\frac{1}{2n(n+1)} \leq \gamma_n - \gamma \leq 0$ .
6. 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 $0.5772156\dots$ ).
7. 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](#def-b1-series-def), to second order.**

8. 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: $$\sum_{k=1}^{2m} \frac{(-1)^{k-1}}{k}  = \ln 2 - \frac{1}{4m} + O\Bigl(\frac{1}{m^2}\Bigr) .$$
9. Deduce the asymptotic error of the alternating harmonic [series](#def-b1-series-def) 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](#thm-b1-series-alternating) .
10. (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.
11. 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.**

12. Show that the positive part $\sum \frac{1}{2j-1}$ and the negative part $\sum \frac{1}{2j}$ of the alternating harmonic [series](#def-b1-series-def) both diverge — the signature of *conditional* convergence.
13. Prove the general [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) behind question 12: if $\sum u_n$ converges but $\sum \abs{u_n}$ diverges, then the [series](#def-b1-series-def) 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.
14. (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 $\sum_{n > N}\abs{u_n}$)* .
15. (Riemann’s recipe) Let $t \in \R$ . Describe the greedy rearrangement of the alternating harmonic [series](#def-b1-series-def) : 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](#def-b1-series-def) 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](#def-b1-series-def) in blocks: $p$ positive terms (the next odd reciprocals), then $q$ negative terms (the next even reciprocals), and repeat.

16. Check that this is a genuine rearrangement (every term exactly once), and that for $(p, q) = (1, 2)$ it reads $$1 - \frac12 - \frac14 + \frac13 - \frac16 - \frac18 +  \frac15 - \frac1{10} - \frac1{12} + \dots$$
17. (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.
18. 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: $$\ln 2 + \frac12 \ln\frac pq .$$
19. Control the partial sums *inside* a block (the terms tend to $0$ ) and conclude that the $(p,  q)$ -rearranged [series](#def-b1-series-def) 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 $0.3466$ .
20. 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.**

21. Identify the sum of the convergent [series](#def-b1-series-def) $\sum_{k\geq1} \bigl(\frac1k - \ln\frac{k+1}{k}\bigr)$ ( [Exercise 17.7](#exo-b1-series-7) ): show it equals $\gamma$ .
22. 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.
23. Show that some rearrangement of the alternating harmonic [series](#def-b1-series-def) diverges to $+\infty$ *(blocks of positive terms long enough to gain $1$ each time, using question 12, separated by single negative terms)* .
24. Sharpen [Example 17.11](#ex-b1-series-harmonicstack) 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](#pb-b1-series-1) is exactly the correction the crude framing was missing.
25. 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](#thm-b1-series-absolute) 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](#def-b1-series-def) and products of [series](#def-b1-series-def) 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 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](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-adjacent)), with common limit $\lim
a_n = \gamma$ ([Exercise 17.7](#exo-b1-series-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](#ex-b1-series-ln2) 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](#def-b1-series-def) 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](#def-b1-series-def)), 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](#thm-b1-series-absolute) 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](#def-b1-series-def) 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](#def-b1-series-def).

**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](#def-b1-series-def) 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](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-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](#def-b1-series-def) of [Exercise 17.7](#exo-b1-series-7) sums exactly to [Euler’s constant](#pb-b1-series-1).

**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](#ex-b1-series-harmonicstack), and pinned by $\gamma$.

**25.** (i) In $H_n = \ln n + \gamma + \frac{1}{2n} +
O(n^{-2})$: the $\ln n$ is the [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def), $\gamma$ the price of replacing a sum by an [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) (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](#thm-b1-series-absolute) 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](#def-b1-series-def) 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.