---
title: "Sequences"
book: "University Mathematics — Year 1"
subject: math
language: en
chapter: 11
exercises: 12
source: https://one-course.com/books/math/3/en/chapter/11-sequences
---

# Chapter 11 — Sequences

Sequences were manipulated in the High School volume with the limit concept taken half on faith. Here the theory is rebuilt on the completeness of $\R$ ([Chapter 10](https://one-course.com/books/math/3/en/chapter/10-real-numbers#ch-b1-reals)): every classical theorem — monotone convergence, [adjacent sequences](#thm-b1-seq-adjacent), Bolzano–Weierstrass, the Cauchy criterion — is a face of that single axiom. The chapter ends with the practical study of sequences defined by $u_{n+1} =
f(u_n)$.

## 11.1 Convergence

**Definition 11.1 (Limit of a sequence).**

A sequence $(u_n)$ of reals *converges* to $\ell \in \R$ when

$$
\forall \varepsilon > 0,\ \exists N \in \N,\ \forall n \geq N,
\qquad \abs{u_n - \ell} \leq \varepsilon .
$$

One writes $u_n \to \ell$ or $\lim u_n = \ell$. A sequence that does not converge (to any real) *diverges*. Divergence *to $+\infty$*: $\forall M,\ \exists N,\ \forall n \geq N,\ u_n \geq M$ (similarly $-\infty$).

**Example 11.2 (An ε\varepsilonε–NNN proof, written out once).**

Claim: $u_n = \dfrac{n^2 + 1}{2n^2 - 3} \to \dfrac12$. First isolate the error:

$$
\Bigl| u_n - \frac12 \Bigr|
= \Bigl| \frac{2(n^2 + 1) - (2n^2 - 3)}{2(2n^2 - 3)} \Bigr|
= \frac{5}{2\,\abs{2n^2 - 3}}
= \frac{5}{2\,(2n^2 - 3)} \quad (n \geq 2).
$$

Then dominate it by something simple: for $n \geq 2$, $2n^2 - 3
\geq n^2$, so the error is $\leq \frac{5}{2n^2} \leq \frac 5{2n}$. Given $\varepsilon > 0$, the [Archimedean property](https://one-course.com/books/math/3/en/chapter/10-real-numbers#thm-b1-reals-archimedes) provides $N \geq
\max\bigl(2, \frac{5}{2\varepsilon}\bigr)$; for $n \geq N$ the error is $\leq \varepsilon$. Done. The closing insight: an $\varepsilon$–$N$ proof has exactly three moves — compute the error, bound it by a decreasing elementary expression, solve for the threshold — and after this chapter’s theorems (operations, squeeze) one almost never writes such a proof again: the theorems package the three moves once and for all.

**Example 11.3 (Divergence to infinity, certified).**

Claim: $u_n = n^2 - 100n \to +\infty$. Factor the dominant term: $u_n = n^2\bigl(1 - \frac{100}{n}\bigr) \geq \frac{n^2}{2}$ for $n \geq 200$. Given $M$, take $N = \max\bigl(200,
\lceil\sqrt{2M}\rceil\bigr)$: for $n \geq N$, $u_n \geq
\frac{n^2}{2} \geq M$. Two habits are on display: dominant-term factoring converts a competition ($n^2$ against $-100n$) into a single scale times a factor tending to $1$; and the threshold may be huge ($u_{100} = 0$, the sequence is even negative before $n = 100$) — divergence to $+\infty$ is a [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) about the tail, indifferent to any finite amount of misbehavior.

**Proposition 11.4 (First properties).**

1. The limit, if it exists, is unique.
2. A convergent sequence is bounded.
3. If $u_n \to \ell$ , every modification of finitely many terms leaves the convergence and the limit unchanged.

**Proof.** (1) If $u_n \to \ell$ and $u_n \to \ell'$ with $\ell \neq \ell'$, take $\varepsilon = \frac{\abs{\ell - \ell'}}{3}$: beyond the two thresholds, $\abs{\ell - \ell'} \leq \abs{\ell - u_n} + \abs{u_n -
\ell'} \leq 2\varepsilon = \frac23 \abs{\ell - \ell'}$, absurd.

(2) With $\varepsilon = 1$: beyond $N$, $\abs{u_n} \leq \abs\ell +
1$; the finitely many earlier terms are bounded too, so $\abs{u_n}
\leq \max(\abs{u_0}, \dots, \abs{u_{N-1}}, \abs\ell + 1)$.

(3) In detail: suppose $v_n = u_n$ for $n \geq n_0$ and $u_n \to
\ell$. Given $\varepsilon > 0$, take the threshold $N$ for $(u_n)$: for $n \geq \max(N, n_0)$, $\abs{v_n - \ell} = \abs{u_n
- \ell} \leq \varepsilon$. So $v_n \to \ell$: the definition quantifies only over $n \geq N$, and any finite prefix can be overwritten at the cost of enlarging the threshold. (This is why hypotheses “for all large $n$” suffice everywhere in this chapter.) ∎

**Theorem 11.5 (Operations on limits).**

If $u_n \to \ell$ and $v_n \to m$, then

$$
u_n + v_n \to \ell + m, \qquad u_n v_n \to \ell m, \qquad
\frac{u_n}{v_n} \to \frac{\ell}{m} \ (\text{if } m \neq 0),
\qquad \abs{u_n} \to \abs\ell .
$$

**Proof.** *Sum:* $\abs{(u_n + v_n) - (\ell + m)} \leq \abs{u_n - \ell} +
\abs{v_n - m} \leq 2\varepsilon$ beyond the larger threshold. *Product:* write

$$
u_n v_n - \ell m = (u_n - \ell)\,v_n + \ell\,(v_n - m);
$$

$(v_n)$ is bounded by some $B$ ([Proposition 11.4](#prop-b1-seq-first)), so the right side is $\leq B\abs{u_n - \ell} + \abs{\ell}\,\abs{v_n - m}$, arbitrarily small. *Quotient:* it suffices to treat $\frac
1{v_n}$. With $\varepsilon = \frac{\abs m}{2}$: beyond some $N_0$, $\abs{v_n} \geq \frac{\abs m}{2}$, so

$$
\Bigl| \frac{1}{v_n} - \frac 1m \Bigr|
= \frac{\abs{m - v_n}}{\abs{v_n m}}
\leq \frac{2}{m^2}\,\abs{v_n - m} \longrightarrow 0 .
$$

*Absolute value:* $\bigl|\abs{u_n} - \abs\ell\bigr| \leq
\abs{u_n - \ell}$ (reverse triangle inequality, [Proposition 3.2](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#prop-b1-complex-rules)). ∎

**Example 11.6 (Operations plus one algebraic trick).**

Compute $\lim\,\bigl(\sqrt{n^2 + n} - n\bigr)$. The two pieces separately tend to $+\infty$: the operations theorem says nothing about their difference (an *indeterminate form*). Multiply by the conjugate:

$$
\sqrt{n^2 + n} - n
= \frac{(n^2 + n) - n^2}{\sqrt{n^2 + n} + n}
= \frac{n}{\sqrt{n^2+n} + n}
= \frac{1}{\sqrt{1 + \frac1n} + 1} .
$$

Now everything converges: $\sqrt{1 + \frac1n} \to 1$, because $0 \leq \sqrt{1 + h} - 1 = \frac{h}{\sqrt{1+h} + 1} \leq h$ (conjugate again, then squeeze with $h = \frac1n$); then the operations theorem gives the limit $\frac{1}{1 + 1} =
\frac12$. The closing insight: the operations theorem is not a calculator for all limits — indeterminate forms ($\infty -
\infty$, $\frac00$, $0 \times \infty$, $1^\infty$) must first be *transformed* by algebra (conjugates, factoring the dominant term) until every piece converges; the systematic machine for resistant cases is the asymptotic expansion of [Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor).

**Theorem 11.7 (Limits and order).**

1. If $u_n \leq v_n$ for all large $n$ , and both converge, then $\lim u_n \leq \lim v_n$ . (Strict inequalities do *not* pass to the limit: $\frac 1n > 0$ but $\lim = 0$ .)
2. (Squeeze theorem) If $u_n \leq w_n  \leq v_n$ for all large $n$ and $u_n, v_n \to \ell$ , then $w_n \to \ell$ .
3. If $u_n \to \ell > 0$ , then $u_n > \frac\ell2 > 0$ for all large $n$ .

**Proof.** (1) Suppose $\ell = \lim u_n > m = \lim v_n$; with $\varepsilon =
\frac{\ell - m}{3}$, large terms satisfy $v_n \leq m + \varepsilon <
\ell - \varepsilon \leq u_n$, contradicting $u_n \leq v_n$.

(2) Beyond the thresholds: $\ell - \varepsilon \leq u_n \leq w_n \leq
v_n \leq \ell + \varepsilon$.

(3) is [Definition 11.1](#def-b1-seq-limit) with $\varepsilon = \frac\ell2$. ∎

**Example 11.8 (Two squeezes).**

(i) $\dfrac{\sin n}{n} \to 0$: from $-\frac1n \leq \frac{\sin
n}{n} \leq \frac1n$, both walls collapsing on $0$ — no need to understand the erratic numerator at all. (ii) $(2^n +
3^n)^{1/n} \to 3$: frame the inside,

$$
3^n \leq 2^n + 3^n \leq 2\cdot3^n
\quad\Longrightarrow\quad
3 \leq (2^n + 3^n)^{1/n} \leq 3\cdot 2^{1/n} ,
$$

and $2^{1/n} = \eu^{\frac{\ln 2}{n}} \to 1$ (as for $5^{1/n}$ in [Exercise 11.2](#exo-b1-seq-2)): the squeeze delivers $3$. The closing insight: a sum of competing exponentials behaves like its *largest* term — the smaller ones are absorbed by a harmless constant factor, which the $n$-th root then erases.

## 11.2 Monotone sequences

**Theorem 11.9 (Monotone limit theorem).**

An increasing sequence bounded above converges, to $\sup\{u_n : n \in
\N\}$; an increasing sequence not bounded above diverges to $+\infty$. (Mirror [statements](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) for decreasing sequences.)

**Proof.** Let $s = \sup\{u_n\}$ ([Theorem 10.2](https://one-course.com/books/math/3/en/chapter/10-real-numbers#thm-b1-reals-sup)). Given $\varepsilon >
0$, the $\varepsilon$-characterization ([Proposition 10.4](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-epsilon)) yields $N$ with $u_N > s -
\varepsilon$; by monotonicity, $s - \varepsilon < u_N \leq u_n \leq
s$ for all $n \geq N$: convergence to $s$. If unbounded: for every $M$ some $u_N > M$, and monotonicity keeps all later terms above $M$. ∎

**Example 11.10 (The monotone theorem as an existence machine).**

Let $u_n = \prod_{k=1}^{n} \bigl(1 + \frac{1}{2^k}\bigr)$. Each factor exceeds $1$, so $(u_n)$ is increasing. Bounded above? Take logarithms and use $\ln(1 + x) \leq x$ ([Example 14.20](https://one-course.com/books/math/3/en/chapter/14-differentiation#ex-b1-derivative-convexineq) anticipates it; or the crude $1 + x \leq \eu^x$ from the High School volume):

$$
\ln u_n = \sum_{k=1}^{n} \ln\Bigl(1 + \frac{1}{2^k}\Bigr)
\leq \sum_{k=1}^{n} \frac{1}{2^k} < 1 ,
$$

so $u_n < \eu$. Increasing and bounded: $(u_n)$ converges to some $\ell \in \intoc{u_1}{\eu}$ — a perfectly well-defined real number with no closed form in sight ($\ell = 2.384\dots$). The closing insight: the monotone limit theorem is the cheapest existence machine in analysis; it named $\eu$ itself ([Example 11.12](#ex-b1-seq-e) below), and in [Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series) it will decide the convergence of every positive series by mere boundedness.

**Theorem 11.11 (Adjacent sequences).**

Let $(a_n)$ be increasing, $(b_n)$ decreasing, with $b_n - a_n \to
0$. Then both converge, to a *common* limit $\ell$, and $a_n
\leq \ell \leq b_n$ for all $n$.

**Proof.** First, $a_n \leq b_n$ for all $n$: the sequence $(b_n - a_n)$ is decreasing and tends to $0$, so it is $\geq 0$ (a negative term would freeze it below $0$). Then $(a_n)$ is increasing and bounded above by $b_0$: it converges to some $\ell$ ([Theorem 11.9](#thm-b1-seq-monotone)); likewise $(b_n) \to \ell'$; and $\ell' - \ell = \lim (b_n - a_n) =
0$. The inequalities $a_n \leq \ell \leq b_n$ follow from monotonicity ($\ell = \sup a_k \geq a_n$, etc.). ∎

![Adjacent sequences: (a_n) climbs, (b_n) descends, and the gap between them shrinks to 0. Each interval (a_n, b_n) contains all later ones, and the common limit is the unique point left in every interval — the picture behind the dichotomy proofs of Bolzano–Weierstrass below and of the intermediate value theorem in .](https://one-course.com/images/onecourse/chapters/math-3/b1-seq/fig-3a699c2487eb.svg)

*[Adjacent sequences](#thm-b1-seq-adjacent): $(a_n)$ climbs, $(b_n)$ descends, and the gap between them shrinks to $0$. Each [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) $\intcc{a_n}{b_n}$ contains all later ones, and the common limit $\ell$ is the unique point left in every [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) — the picture behind the dichotomy proofs of Bolzano–Weierstrass below and of the intermediate value theorem in [Chapter 13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#ch-b1-continuity).*

**Example 11.12 (The number e\eue).**

Set $a_n = \sum_{k=0}^{n} \frac{1}{k!}$ and $b_n = a_n + \frac{1}{n
\cdot n!}$ ($n \geq 1$). Then $(a_n)$ increases; and

$$
b_{n+1} - b_n = \frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} -
\frac{1}{n\,n!}
= \frac{n(n+1) + n - (n+1)^2}{n(n+1)(n+1)!}
= \frac{-1}{n(n+1)(n+1)!} < 0 ,
$$

so $(b_n)$ decreases, and $b_n - a_n \to 0$: adjacent. Their common limit is (by definition here) the number $\eu \approx 2.71828$; the inequalities $a_n < \eu < b_n$ are sharp enough to prove $\eu \notin
\Q$ ([Exercise 11.9](#exo-b1-seq-9)).

## 11.3 Subsequences and Bolzano–Weierstrass

**Definition 11.13 (Subsequence).**

A *subsequence* of $(u_n)$ is a sequence $(u_{\varphi(n)})$ where $\varphi \colon \N \to \N$ is strictly increasing (note $\varphi(n) \geq n$, by induction).

**Proposition 11.14.**

If $u_n \to \ell$ ($\ell \in \R$ or $\pm\infty$), every [subsequence](#def-b1-seq-subsequence) tends to $\ell$. Consequently, a sequence with two [subsequences](#def-b1-seq-subsequence) of different limits diverges. Conversely, if $(u_{2n})$ and $(u_{2n+1})$ both converge to the *same* $\ell$, then $u_n \to
\ell$.

**Proof.** Beyond the threshold $N$ for $(u_n)$, all indices $\varphi(n) \geq n
\geq N$ qualify (the inequality $\varphi(n) \geq n$ is the induction noted in [Definition 11.13](#def-b1-seq-subsequence): $\varphi(0)
\geq 0$, and $\varphi(n+1) > \varphi(n) \geq n$ forces $\varphi(n+1) \geq n + 1$). For the converse: given $\varepsilon$, take the two thresholds $N_0$ (evens) and $N_1$ (odds); an arbitrary index $n \geq \max(2N_0, 2N_1 + 1)$ is either even, $n = 2k$ with $k \geq N_0$, or odd, $n = 2k+1$ with $k \geq N_1$ — in both cases $\abs{u_n - \ell} \leq
\varepsilon$: every index is covered by one of the two [subsequences](#def-b1-seq-subsequence), and that is the whole point. ∎

**Example 11.15 (Subsequential limits).**

For $u_n = (-1)^n \frac{n}{n+1}$: the even [subsequence](#def-b1-seq-subsequence) tends to $1$, the odd one to $-1$, so the sequence diverges — but it does so in an organized way, clustering around the two values $\pm 1$. For $u_n = \cos\frac{2\pi n}{3}$: the three [subsequences](#def-b1-seq-subsequence) of indices $3k$, $3k + 1$, $3k + 2$ are constant, equal to $1$, $-\frac12$, $-\frac12$; the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of subsequential limits is $\{1,
-\frac12\}$. The closing insight: a *bounded* sequence converges exactly when it has a single subsequential limit ([Exercise 11.8](#exo-b1-seq-8)); divergence of a bounded sequence always means at least two clusters, and Bolzano–Weierstrass below guarantees there is at least one.

**Theorem 11.16 (Bolzano–Weierstrass).**

Every bounded sequence of reals has a convergent [subsequence](#def-b1-seq-subsequence).

**Proof.** Let $u_n \in \intcc{a}{b}$ for all $n$. Build nested segments by dichotomy: set $\intcc{a_0}{b_0} = \intcc{a}{b}$; given $\intcc{a_k}{b_k}$ containing $u_n$ for infinitely many $n$, one of its two halves still contains $u_n$ for infinitely many $n$ — call it $\intcc{a_{k+1}}{b_{k+1}}$. The sequences $(a_k)$, $(b_k)$ are adjacent ($b_k - a_k = \frac{b-a}{2^k} \to 0$), with common limit $\ell$ ([Theorem 11.11](#thm-b1-seq-adjacent)).

Extract: choose $\varphi(0)$ with $u_{\varphi(0)} \in
\intcc{a_0}{b_0}$, then, inductively, $\varphi(k+1) > \varphi(k)$ with $u_{\varphi(k+1)} \in \intcc{a_{k+1}}{b_{k+1}}$ — possible since that segment contains infinitely many terms. Then $a_k \leq
u_{\varphi(k)} \leq b_k$, and the squeeze theorem gives $u_{\varphi(k)} \to \ell$. ∎

**Remark 11.17 (What Bolzano–Weierstrass does, and does not, say).**

It *does* say: from boundedness alone, some [subsequence](#def-b1-seq-subsequence) converges — existence with no formula, as the dichotomy proof makes clear (nothing tells us *which* indices survive). It does *not* say the limit is unique: $((-1)^n)$ has [subsequences](#def-b1-seq-subsequence) converging to $1$ and to $-1$, and the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of subsequential limits can even be infinite ([Example 11.15](#ex-b1-seq-sublimits), and all of the Cantor [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) in [Problem 12.1](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#pb-b1-topology-1)). It does not survive unboundedness: $(n)$ has no convergent [subsequence](#def-b1-seq-subsequence) at all — though one can always extract a [subsequence](#def-b1-seq-subsequence) tending to $+\infty$ or $-\infty$ from any unbounded sequence (choose $\varphi(k)$ with $u_{\varphi(k)} \geq k$, say). Used correctly, the theorem is an *existence pump*: it appears at the crux of the Cauchy criterion below, of Heine’s theorem, and of the extreme value theorem — always to produce a point that no explicit construction offers.

## 11.4 Cauchy sequences and completeness

**Definition 11.18.**

A sequence $(u_n)$ is a *Cauchy sequence* when its terms become arbitrarily close *to each other*:

$$
\forall \varepsilon > 0,\ \exists N,\ \forall p, q \geq N,
\qquad \abs{u_p - u_q} \leq \varepsilon .
$$

**Example 11.19 (Verifying the Cauchy property by hand).**

Let $u_n = \sum_{k=0}^{n} \frac{\cos k}{2^k}$ — no monotonicity, no guessable limit. For $p > q$:

$$
\abs{u_p - u_q}
= \Bigl| \sum_{k=q+1}^{p} \frac{\cos k}{2^k} \Bigr|
\leq \sum_{k=q+1}^{p} \frac{1}{2^k}
< \frac{1}{2^{q}} ,
$$

by the triangle inequality, $\abs{\cos k} \leq 1$ and a finite geometric sum. Given $\varepsilon > 0$, choose $N$ with $2^{-N}
\leq \varepsilon$: all gaps beyond $N$ are $\leq \varepsilon$, the sequence is Cauchy, hence converges — to a limit nobody can name in closed form, which is exactly the point. The closing insight: geometric domination of the increments is the standard way to earn the Cauchy property, and [Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series) will bottle the argument as “absolute convergence implies convergence”.

**Theorem 11.20 (Completeness of R\RR).**

A sequence of reals converges if and only if it is a [Cauchy sequence](#def-b1-seq-cauchy).

**Proof.** ($\Rightarrow$) If $u_n \to \ell$: beyond the threshold for $\frac\varepsilon2$, $\abs{u_p - u_q} \leq \abs{u_p - \ell} +
\abs{\ell - u_q} \leq \varepsilon$.

($\Leftarrow$) Let $(u_n)$ be Cauchy. *It is bounded*: with $\varepsilon = 1$, beyond $N$ all terms lie within $1$ of $u_N$, and the head is finite. *Extract*: by [Theorem 11.16](#thm-b1-seq-bw), some [subsequence](#def-b1-seq-subsequence) $u_{\varphi(n)} \to \ell$. *Conclude*: given $\varepsilon > 0$, take $N$ (Cauchy, for $\frac\varepsilon2$) and $n \geq N$ with $\abs{u_{\varphi(n)} - \ell}
\leq \frac\varepsilon2$ and $\varphi(n) \geq N$; then for every $p
\geq N$:

$$
\abs{u_p - \ell} \leq \abs{u_p - u_{\varphi(n)}} +
\abs{u_{\varphi(n)} - \ell} \leq \varepsilon . \qedhere
$$

∎

**Remark 11.21.**

The value of the criterion: it certifies convergence *without naming the limit*. It fails over $\Q$ (the decimal truncations of $\sqrt 2$ form a [Cauchy sequence](#def-b1-seq-cauchy) of rationals with no rational limit): completeness is a property of $\R$, equivalent to the [upper bound](https://one-course.com/books/math/3/en/chapter/10-real-numbers#def-b1-reals-bounds) axiom. It is also the workhorse behind the convergence of series ([Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series)).

**Example 11.22 (A Cauchy sequence with an invisible limit).**

Let $S_n = \sum_{k=1}^{n} \frac{1}{k^2}$. For $p > q \geq 1$:

$$
S_p - S_q = \sum_{k=q+1}^{p} \frac{1}{k^2}
\leq \sum_{k=q+1}^{p} \frac{1}{k(k-1)}
= \sum_{k=q+1}^{p} \Bigl(\frac{1}{k-1} - \frac 1k\Bigr)
= \frac 1q - \frac 1p < \frac 1q ,
$$

so beyond $N > \frac1\varepsilon$ all gaps are $\leq \varepsilon$: $(S_n)$ is Cauchy, hence converges. Notice what just happened: we proved that a specific real number exists without having any name for it. (It is $\frac{\pi^2}{6}$ — a celebrated identity of Euler, proved in the Year 2 volume; nothing in this chapter could tell us that.) This division of labor — existence now, identification later, if ever — is the Cauchy criterion’s whole point, and the engine of the theory of series in [Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series).

## 11.5 Recurrent sequences

**Method 11.23 (Studying un+1=f(un)u_{n+1} = f(u_n)un+1​=f(un​)).**

Given $f$ and a starting point $u_0$:

1. *Stable [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals):* find an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) $I$ with $f(I)  \subseteq I$ containing $u_0$ : then all $u_n \in I$ (by induction).
2. *Candidate limits:* if $u_n \to \ell \in I$ and $f$ is continuous at $\ell$ ( [Chapter 13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#ch-b1-continuity) ), then $\ell$ is a *fixed point* : $f(\ell) = \ell$ . Solve $f(x) = x$ .
3. *Monotonicity:* if $f$ is increasing on $I$ , then $(u_n)$ is monotonic (increasing if $u_1 \geq u_0$ , decreasing otherwise); combined with boundedness, [Theorem 11.9](#thm-b1-seq-monotone) concludes. If $f$ is decreasing, study the two [subsequences](#def-b1-seq-subsequence) $(u_{2n})$ and $(u_{2n+1})$ , which are monotonic for $f \circ f$ .
4. *Error control:* an inequality $\abs{f(x) - \ell} \leq  k\abs{x - \ell}$ with $k < 1$ gives $\abs{u_n - \ell} \leq  k^n \abs{u_0 - \ell} \to 0$ directly.

**Example 11.24 (Heron’s method).**

Let $u_0 = 2$ and $u_{n+1} = \dfrac12\Bigl(u_n +
\dfrac{2}{u_n}\Bigr)$: the ancient algorithm for $\sqrt 2$.

- *Stability:* for $x > 0$ , the arithmetic–geometric mean inequality gives $\frac12(x + \frac2x) \geq \sqrt{x \cdot  \frac 2x} = \sqrt 2$ ; so $I = \intco{\sqrt 2}{+\infty}$ is stable and contains $u_1$ (indeed $u_1 = \frac32 \geq  \sqrt 2$ ).
- *Monotonicity:* for $x \geq \sqrt 2$ , $\;x - f(x) =  \frac{x^2 - 2}{2x} \geq 0$ : the sequence decreases from $u_1$ on, and is bounded below by $\sqrt 2$ : it converges.
- *Limit:* the fixed points solve $x = \frac12(x +  \frac2x)$ , i.e. $x^2 = 2$ : on $I$ , $\ell = \sqrt 2$ .
- *Speed:* $u_{n+1} - \sqrt 2 = \frac{(u_n -  \sqrt2)^2}{2u_n} \leq \frac{(u_n - \sqrt 2)^2}{2\sqrt 2}$ : the number of correct digits roughly *doubles* at each step (quadratic convergence).

![Heron’s iteration u_n+1 = 1/2 (u_n + 2/u_n ), drawn as a staircase between the graph of f and the diagonal y = x: from u_0 = 2, the iterates slide down to the fixed point √ 2.](https://one-course.com/images/onecourse/chapters/math-3/b1-seq/fig-1827245c9f25.svg)

*Heron’s iteration $u_{n+1} = \frac12\bigl(u_n +
\frac{2}{u_n}\bigr)$, drawn as a staircase between the graph of $f$ and the diagonal $y = x$: from $u_0 = 2$, the iterates slide down to the fixed point $\sqrt 2$.*

**Remark 11.25 (Common pitfalls with limits).**

Four classics. (i) *Small steps do not imply convergence*: $u_{n+1} - u_n \to 0$ is far weaker than the Cauchy property — the harmonic sums $H_n$ have steps $\frac{1}{n+1} \to 0$ yet diverge to $+\infty$ ([Exercise 11.5](#exo-b1-seq-5)); the Cauchy condition controls $\abs{u_p - u_q}$ for *all* large pairs, not consecutive ones. (ii) *Strict inequalities die in the limit*: from $u_n < v_n$ for all $n$ one gets only $\lim u_n
\leq \lim v_n$ ([Theorem 11.7](#thm-b1-seq-order)); $\frac1n > 0$ and yet $\lim = 0$. (iii) *Bounded is not convergent*: $((-1)^n)$ is bounded and diverges; boundedness plus *monotonicity* converges, boundedness alone only guarantees a convergent [subsequence](#def-b1-seq-subsequence) ([Theorem 11.16](#thm-b1-seq-bw)). (iv) *The fixed-point equation comes second, not first*: for $u_{n+1} = f(u_n)$, solving $f(\ell) = \ell$ identifies the limit *only after* convergence is proved. The recurrence $u_{n+1} = 2u_n$ has the unique fixed point $\ell = 0$, yet from $u_0 = 1$ the sequence runs to $+\infty$: the equation $\ell = 2\ell$ was never entitled to a limit. Order of operations, always: existence first ([Method 11.23](#met-b1-seq-recurrent), steps 1–3), identification second.

**Example 11.26 (A decreasing fff: the golden recurrence).**

Let $u_0 = 1$ and $u_{n+1} = \dfrac{1}{1 + u_n}$. Here $f(x) =
\frac{1}{1+x}$ is *decreasing*, so the sequence is not monotonic (it alternates around its limit); the contraction step of [Method 11.23](#met-b1-seq-recurrent) is the right tool. Stability: if $x \in \intcc{\frac12}{1}$ then $1 + x \in
\intcc{\frac32}{2}$, so $f(x) \in \intcc{\frac12}{\frac23}
\subseteq \intcc{\frac12}{1}$, and $u_1 = \frac12$ puts the whole sequence there. Fixed point: $\ell = \frac{1}{1+\ell}$ with $\ell > 0$ gives $\ell^2 + \ell - 1 = 0$, i.e.

$$
\ell = \frac{\sqrt5 - 1}{2} = 0.6180\dots
$$

(the golden ratio’s inverse). Contraction: for $x, y \in
\intcc{\frac12}{1}$,

$$
\abs{f(x) - f(y)} = \frac{\abs{x - y}}{(1+x)(1+y)}
\leq \frac{\abs{x-y}}{(3/2)^2} = \frac49\,\abs{x - y} ,
$$

so $\abs{u_n - \ell} \leq \bigl(\frac49\bigr)^{n-1}\abs{u_1 -
\ell} \to 0$: convergence, with geometric speed, no monotonicity needed. The closing insight: monotone methods and contraction methods split the recurrent world between them — increasing $f$ gives monotone orbits, decreasing $f$ gives alternating orbits tamed by a Lipschitz constant $< 1$ (the systematic theory is [Exercise 14.11](https://one-course.com/books/math/3/en/chapter/14-differentiation#exo-b1-derivative-11)).

**Remark 11.27 (Perspectives inside this volume).**

Sequences are the measuring instrument the rest of the volume holds up to every object. In [Chapter 12](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#ch-b1-topology) they *characterize* closedness and compactness; in [Chapter 13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#ch-b1-continuity) they transport limits of functions; in [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration) Riemann sums are sequences converging to the integral; [Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series) *is* the theory of one special class of sequences, the partial sums. Even the algebra chapters consume them: the iterates of a matrix in [Chapter 21](https://one-course.com/books/math/3/en/chapter/21-matrices#ch-b1-matrices) form sequences whose behavior (convergence of $A^n$) is a linear-algebra question with this chapter’s vocabulary. The two theorems to carry everywhere: monotone limit (existence from order) and Bolzano–Weierstrass (existence from boundedness) — between them, nearly every limit in this book is born.

**Remark 11.28 (Complex sequences).**

A sequence $(z_n)$ of complex numbers converges to $\ell$ when $\abs{z_n - \ell} \to 0$; equivalently, when $\Re(z_n) \to \Re(\ell)$ and $\Im(z_n) \to \Im(\ell)$ (compare $\abs{z}$ with $\abs{\Re z} +
\abs{\Im z}$). The theorems not involving order — operations, Bolzano–Weierstrass (extract twice), Cauchy criterion — carry over verbatim.

## 11.6 Exercises

**Exercise 11.1 ★.**

Directly from [Definition 11.1](#def-b1-seq-limit), prove that $\dfrac{2n+1}{n+3} \to 2$, and that $(u_n) = ((-1)^n)$ diverges.

**Solution of Exercise 11.1.**

$\Bigl|\dfrac{2n+1}{n+3} - 2\Bigr| = \dfrac{5}{n+3}$. Given $\varepsilon > 0$, take $N > \frac 5\varepsilon - 3$ (Archimedes): for $n \geq N$, $\frac{5}{n+3} \leq \varepsilon$. Hence the limit is $2$.

$((-1)^n)$: its [subsequences](#def-b1-seq-subsequence) $(u_{2n}) = (1)$ and $(u_{2n+1}) =
(-1)$ converge to different limits, so the sequence diverges ([Proposition 11.14](#prop-b1-seq-subsequences)). (Directly: any candidate $\ell$ fails for $\varepsilon = \frac12$, since consecutive terms are at distance $2$.)

**Exercise 11.2 ★.**

Compute the limits:

$$
\frac{n^2 - 3n + 1}{2n^2 + 5},
\qquad
\sqrt{n+1} - \sqrt n,
\qquad
\frac{2^n + n^3}{3^n - n^2},
\qquad
\sqrt[n]{5}\ \Bigl(= 5^{1/n}\Bigr).
$$

**Solution of Exercise 11.2.**

Dividing by $n^2$: $\dfrac{1 - 3/n + 1/n^2}{2 + 5/n^2} \to
\dfrac12$.

$\sqrt{n+1} - \sqrt n = \dfrac{1}{\sqrt{n+1} + \sqrt n} \to 0$ (conjugate).

$\dfrac{2^n + n^3}{3^n - n^2} = \dfrac{(2/3)^n + n^3/3^n}{1 -
n^2/3^n} \to \dfrac{0 + 0}{1 - 0} = 0$, using $q^n \to 0$ for $\abs q
< 1$ and the polynomial-vs-geometric comparison ([Proposition 4.6](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-powerrules)).

$5^{1/n} = \eu^{\frac{\ln 5}{n}} \to \eu^0 = 1$.

**Exercise 11.3 ★.**

Prove the standard comparison: if $\abs{q} < 1$ then $q^n \to 0$ *(write $\frac{1}{\abs q} = 1 + h$, $h > 0$, and use the Bernoulli inequality $(1+h)^n \geq 1 + nh$, to be proved by induction)*. What are the behaviors for $q = 1$, $q = -1$, $\abs q >
1$?

**Solution of Exercise 11.3.**

Bernoulli: $(1+h)^n \geq 1 + nh$ for $h \geq -1$, by induction — $(1+h)^{n+1} = (1+h)^n(1+h) \geq (1+nh)(1+h) = 1 + (n+1)h + nh^2
\geq 1 + (n+1)h$.

For $0 < \abs q < 1$: write $\frac{1}{\abs q} = 1 + h$, $h > 0$; then $\abs{q}^n = \frac{1}{(1+h)^n} \leq \frac{1}{1 + nh} \to 0$, and the squeeze gives $q^n \to 0$ (the case $q = 0$ is trivial). For $q = 1$: constant sequence, limit $1$. For $q = -1$: diverges ([Exercise 11.1](#exo-b1-seq-1)). For $\abs q > 1$: $\abs q^n = (1 + h)^n \geq
1 + nh \to +\infty$, so $(q^n)$ is unbounded, hence divergent (to $+\infty$ if $q > 1$; with alternating signs, no limit, if $q <
-1$).

**Exercise 11.4 ★.**

Let $u_{n+1} = \frac{u_n + 3}{2}$, $u_0 = 0$. Find the fixed point $\ell$, prove that $v_n = u_n - \ell$ is geometric, and give an explicit formula and the limit of $(u_n)$.

**Solution of Exercise 11.4.**

Fixed point: $\ell = \frac{\ell + 3}{2}$ gives $\ell = 3$. Then

$$
v_{n+1} = u_{n+1} - 3 = \frac{u_n + 3}{2} - 3 = \frac{u_n - 3}{2} =
\frac{v_n}{2}:
$$

$(v_n)$ is geometric with ratio $\frac12$, $v_0 = -3$. So $u_n = 3 -
\frac{3}{2^n} \to 3$.

**Exercise 11.5 ★★.**

(Harmonic series) Let $H_n = \sum_{k=1}^{n} \frac 1k$. Prove that $H_{2n} - H_n \geq \frac12$ for all $n \geq 1$, and conclude that $(H_n)$ is *not* a [Cauchy sequence](#def-b1-seq-cauchy), hence diverges (to $+\infty$, being increasing).

**Solution of Exercise 11.5.**

$H_{2n} - H_n = \sum_{k=n+1}^{2n} \frac 1k \geq n \cdot
\frac{1}{2n} = \frac12$ (each of the $n$ terms is $\geq
\frac{1}{2n}$). If $(H_n)$ were Cauchy, taking $\varepsilon =
\frac13$ would force $\abs{H_{2n} - H_n} \leq \frac13$ for large $n$: contradiction. An increasing non-convergent sequence diverges to $+\infty$ ([Theorem 11.9](#thm-b1-seq-monotone)): $H_n \to +\infty$.

**Exercise 11.6 ★★.**

Suppose $(u_{2n})$, $(u_{2n+1})$ and $(u_{3n})$ all converge. Prove that $(u_n)$ converges. *(Find common [subsequences](#def-b1-seq-subsequence) to equate the limits.)*

**Solution of Exercise 11.6.**

Let $a = \lim u_{2n}$, $b = \lim u_{2n+1}$, $c = \lim u_{3n}$. The sequence $(u_{6n})$ is a [subsequence](#def-b1-seq-subsequence) both of $(u_{2n})$ and of $(u_{3n})$: its limit equals $a$ and $c$, so $a = c$. The sequence $(u_{6n+3})$ is a [subsequence](#def-b1-seq-subsequence) of $(u_{2n+1})$ (odd indices) and of $(u_{3n})$ (indices $6n + 3 = 3(2n+1)$): so $b = c$. Hence $a = b$, and [Proposition 11.14](#prop-b1-seq-subsequences) (evens and odds with equal limits) gives the convergence of $(u_n)$.

**Exercise 11.7 ★★.**

Study the sequence $u_0 = 0$, $u_{n+1} = \sqrt{2 + u_n}$: stability, monotonicity, limit. Then prove the error bound $\abs{u_n - 2} \leq \dfrac{2}{3^{\,n}}$ *(show $2 - u_{n+1} =
\dfrac{2 - u_n}{2 + \sqrt{2 + u_n}}$ and bound the denominator below by $3$)*.

**Solution of Exercise 11.7.**

*Stability and bounds:* $I = \intcc{0}{2}$ is stable: for $x \in
I$, $\sqrt{2 + x} \in \intcc{\sqrt 2}{2} \subseteq I$; and $u_0 = 0
\in I$.

*Monotonicity:* $f(x) = \sqrt{2+x}$ is increasing and $u_1 =
\sqrt 2 > u_0$: by induction $(u_n)$ is increasing. Increasing and bounded above by $2$: it converges ([Theorem 11.9](#thm-b1-seq-monotone)).

*Limit:* $\ell = \sqrt{2 + \ell}$ with $\ell \geq 0$ gives $\ell^2 - \ell - 2 = 0$, so $\ell = 2$.

*Error bound:* multiplying by the conjugate,

$$
2 - u_{n+1} = 2 - \sqrt{2 + u_n}
= \frac{4 - (2 + u_n)}{2 + \sqrt{2+u_n}}
= \frac{2 - u_n}{2 + \sqrt{2 + u_n}}
\leq \frac{2 - u_n}{3},
$$

since $\sqrt{2 + u_n} \geq \sqrt 2 > 1$. By induction from $2 - u_0
= 2$: $\;0 \leq 2 - u_n \leq \frac{2}{3^n}$.

**Exercise 11.8 ★★.**

Let $(u_n)$ be bounded, such that every convergent [subsequence](#def-b1-seq-subsequence) of $(u_n)$ has the *same* limit $\ell$. Prove $u_n \to \ell$. *(Contradiction plus Bolzano–Weierstrass.)*

**Solution of Exercise 11.8.**

Suppose $(u_n)$ does not converge to $\ell$: for some $\varepsilon_0
> 0$, infinitely many indices satisfy $\abs{u_n - \ell} >
\varepsilon_0$; they form a [subsequence](#def-b1-seq-subsequence) $(u_{\varphi(n)})$. This [subsequence](#def-b1-seq-subsequence) is bounded, so by Bolzano–Weierstrass ([Theorem 11.16](#thm-b1-seq-bw)) it has a convergent [sub-subsequence](#def-b1-seq-subsequence), whose limit $\ell'$ satisfies $\abs{\ell' - \ell} \geq \varepsilon_0$ (pass the inequality to the limit, [Theorem 11.7](#thm-b1-seq-order)). But a [sub-subsequence](#def-b1-seq-subsequence) of $(u_n)$ is a convergent [subsequence](#def-b1-seq-subsequence) of $(u_n)$, so by hypothesis $\ell' = \ell$: contradiction.

**Exercise 11.9 ★★★.**

With the notation of [Example 11.12](#ex-b1-seq-e), suppose $\eu = \frac pq$ with $p, q \in \N^*$. Using $a_q < \eu < b_q = a_q + \frac{1}{q\,
q!}$, multiply by $q!$ and derive a contradiction between two integers. Conclude: $\eu$ is irrational.

**Solution of Exercise 11.9.**

Suppose $\eu = \frac pq$, $q \geq 1$. The strict inequalities $a_q <
\eu < a_q + \frac{1}{q\,q!}$ (strict since $(a_n)$ is strictly increasing and $(b_n)$ strictly decreasing) multiplied by $q!$ give

$$
q!\,a_q \;<\; q!\,\frac pq \;<\; q!\,a_q + \frac 1q \leq q!\,a_q + 1.
$$

Now $N = q!\,a_q = \sum_{k=0}^{q} \frac{q!}{k!}$ is an integer (each $\frac{q!}{k!}$ is a product of integers for $k \leq q$), and so is $q!\,\frac pq = (q-1)!\,p$. The display thus places the integer $(q-1)!\,p$ strictly between $N$ and $N + \frac 1q \leq N + 1$: an integer strictly inside $\intoo{N}{N+1}$, which is impossible. Hence $\eu \notin \Q$.

**Exercise 11.10 ★★★.**

(Cesàro means) For a sequence $(u_n)_{n \geq 1}$, set $c_n =
\frac{u_1 + \dots + u_n}{n}$.

1. Prove that $u_n \to \ell$ implies $c_n \to \ell$ *(cut the sum at a threshold $N$; bound the head by a fixed quantity over $n$, the tail by $\varepsilon$)* .
2. Show by example that the converse fails.
3. Deduce that if $u_{n+1} - u_n \to \ell$ , then $\frac{u_n}{n}  \to \ell$ .

**Solution of Exercise 11.10.**

1. Let $\varepsilon > 0$ and $N$ with $\abs{u_k - \ell} \leq  \frac{\varepsilon}{2}$ for $k > N$. For $n > N$: $$\abs{c_n - \ell}  = \Bigl|\frac{\sum_{k=1}^{n}(u_k - \ell)}{n}\Bigr|  \leq \frac{\sum_{k=1}^{N} \abs{u_k - \ell}}{n}  + \frac{n - N}{n}\cdot\frac{\varepsilon}{2}  \leq \frac{C}{n} + \frac{\varepsilon}{2},$$ where $C = \sum_{k=1}^N \abs{u_k - \ell}$ is fixed. For $n$ large, $\frac Cn \leq \frac\varepsilon2$: then $\abs{c_n -  \ell} \leq \varepsilon$.
2. $u_n = (-1)^n$ : diverges, yet $c_n \to 0$ (partial sums bounded by $1$ , divided by $n$ ).
3. Apply (1) to the sequence $v_n = u_{n+1} - u_n \to \ell$ : its Cesàro mean is $\frac{u_{n+1} - u_1}{n} \to \ell$ (telescoping), and $\frac{u_{n+1}}{n} = \frac{u_{n+1} -  u_1}{n} + \frac{u_1}{n} \to \ell$ ; renormalizing indices ( $\frac{u_n}{n} = \frac{u_n}{n-1}\cdot\frac{n-1}{n}$ ) gives $\frac{u_n}{n} \to \ell$ .

**Exercise 11.11 ★★★.**

Let $(u_n)$ satisfy $0 \leq u_{m+n} \leq u_m + u_n$ for all $m, n$ (subadditivity). Prove that $\bigl(\frac{u_n}{n}\bigr)$ converges to $\inf_{n \geq 1} \frac{u_n}{n}$. *(For fixed $m$, write $n = qm
+ r$ and bound $\frac{u_n}{n}$ using $u_n \leq q\,u_m + u_r$.)*

**Solution of Exercise 11.11.**

Let $L = \inf_{n \geq 1} \frac{u_n}{n} \geq 0$, and $\varepsilon >
0$. Choose $m$ with $\frac{u_m}{m} \leq L + \varepsilon$. Every $n$ writes $n = qm + r$, $0 \leq r < m$; subadditivity (iterated) gives $u_n \leq q\,u_m + u_r$, so

$$
\frac{u_n}{n} \leq \frac{q m}{n}\cdot\frac{u_m}{m} + \frac{u_r}{n}
\leq \frac{u_m}{m} + \frac{\max(u_0, \dots, u_{m-1})}{n}
\leq L + \varepsilon + \frac{C_m}{n},
$$

using $qm \leq n$. For $n$ large, $\frac{C_m}{n} \leq \varepsilon$: thus $L \leq \frac{u_n}{n} \leq L + 2\varepsilon$ for all large $n$, which is the convergence to $L$.

**Exercise 11.12 ★★★.**

Using the density of the [subgroup](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-subgroup) $\Z + 2\pi\Z$ of $(\R, +)$ ([Exercise 10.9](https://one-course.com/books/math/3/en/chapter/10-real-numbers#exo-b1-reals-9)), prove that the sequence $(\sin n)_{n \in
\N}$ is dense in $\intcc{-1}{1}$ — in particular it diverges.

**Solution of Exercise 11.12.**

The [subgroup](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-subgroup) $G = \Z + 2\pi\Z$ of $(\R, +)$ is dense: it is not $\alpha\Z$, since $1 = p\alpha$, $2\pi = q\alpha$ would make $2\pi =
\frac qp$ rational — and $\pi \notin \Q$ (admitted here; a proof is sketched in [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration)). By [Exercise 10.9](https://one-course.com/books/math/3/en/chapter/10-real-numbers#exo-b1-reals-9), $G$ is dense in $\R$.

Let now $y \in \intcc{-1}{1}$ and $\theta = \arcsin y$. By density, for every $\varepsilon > 0$ there are $n \in \Z$, $k \in \Z$ with $\abs{(n + 2\pi k) - \theta} \leq \varepsilon$, i.e. $n$ is within $\varepsilon$ of $\theta - 2\pi k$; then, $\sin$ being $2\pi$-periodic and $1$-Lipschitz ($\abs{\sin a - \sin b} \leq
\abs{a - b}$, from the mean value inequality of [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative)),

$$
\abs{\sin n - y} = \abs{\sin(n + 2\pi k) - \sin\theta}
\leq \abs{n + 2\pi k - \theta} \leq \varepsilon .
$$

One detail: $n$ ranges over $\Z$, but $\sin(-n) = -\sin n$ and $y$ was arbitrary in $\intcc{-1}{1}$, so nonnegative indices suffice (replace $(n, y)$ by $(-n, -y)$ if needed). Hence $\{\sin n : n \in
\N\}$ is dense in $\intcc{-1}{1}$; a dense-in-a-segment sequence has [subsequences](#def-b1-seq-subsequence) approaching distinct values, so it diverges.

## 11.7 Problem: Cesàro, Stolz, and the slow fall of the sine

**Problem 11.1.**

Weekend problem — the Cesàro–Stolz theorem and the asymptotics $u_n \sim \sqrt{3/n}$ for $u_{n+1} = \sin
u_n$

The Cesàro–Stolz theorem is the discrete l’Hospital rule: to find the limit of a quotient $a_n/b_n$, it suffices to find the limit of the quotient of *increments* $(a_{n+1} -
a_n)/(b_{n+1} - b_n)$. This problem proves the theorem, harvests classical limits with it, and then aims it at a famous target: the sequence $u_{n+1} = \sin u_n$, which creeps to $0$ at the exactly computable speed $u_n \sim \sqrt{3/n}$. Two facts from the High School volume are granted here and re-proved honestly later in this volume: the tangent-line inequality

$$
\tag{G1} \eu^{u} \geq 1 + u \quad (u \in \R),
$$

re-proved by convexity in [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative), and the sine bracketing

$$
\tag{G2} x - \frac{x^3}{6} \;\leq\; \sin x \;\leq\; x -
\frac{x^3}{6} + \frac{x^5}{120} \quad (0 \leq x \leq 1), \qquad
\abs{\sin x} \leq \abs{x} \quad (x \in \R),
$$

re-proved by Taylor’s formula in [Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor).

**Part I — Sums without closed formulas.**

1. Using $1 + 2 + \dots + n = \frac{n(n+1)}{2}$ and $1^2 +  \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$ , compute $\lim \frac{1 + 2 + \dots + n}{n^2}$ and $\lim \frac{1^2 +  \dots + n^2}{n^3}$ .
2. Let $T_n = \sum_{k=1}^n \sqrt k$, for which no closed formula exists. Prove the bracketing $$\frac{1}{2\sqrt 2}\,n^{3/2} \;\leq\; T_n \;\leq\;  n^{3/2}$$ *(keep only the terms $k > \frac n2$ for the lower bound)*. So $T_n$ has the order $n^{3/2}$ — but which constant? Hold the question until question 8.
3. (Telescoping lemma) Let $(b_n)$ be strictly increasing and suppose that for all $k \geq N$, $$m \;\leq\; \frac{a_{k+1} - a_k}{b_{k+1} - b_k} \;\leq\;  M .$$ Prove that $m \leq \dfrac{a_n - a_N}{b_n - b_N} \leq M$ for every $n > N$.

**Part II — The Cesàro–Stolz theorem.** Let $(b_n)$ be strictly increasing with $b_n \to +\infty$, and suppose $\dfrac{a_{n+1} - a_n}{b_{n+1} - b_n} \to \ell \in \R$.

4. Fix $\varepsilon > 0$ . Show there is $N$ such that $\ell - \varepsilon \leq \dfrac{a_n - a_N}{b_n - b_N}  \leq \ell + \varepsilon$ for all $n > N$ .
5. Establish, for $n > N$, the identity $$\frac{a_n}{b_n} - \ell  = \frac{a_N - \ell\,b_N}{b_n}  + \Bigl(1 - \frac{b_N}{b_n}\Bigr)  \Bigl(\frac{a_n - a_N}{b_n - b_N} - \ell\Bigr),$$ and conclude the theorem: $\dfrac{a_n}{b_n} \to \ell$.
6. Prove the $+\infty$ variant: if $\dfrac{a_{n+1} -  a_n}{b_{n+1} - b_n} \to +\infty$ (same hypotheses on $(b_n)$ ), then $\dfrac{a_n}{b_n} \to +\infty$ .
7. Take $b_n = n$ : recover the Cesàro mean theorem of [Exercise 11.10](#exo-b1-seq-10) . Then show the converse of Cesàro–Stolz fails: for $a_n = (-1)^n$ , $b_n = n$ , the quotient $a_n/b_n$ converges while the quotient of increments does not. Stolz is a one-way street.

**Part III — First dividends.**

8. Prove $(1+h)^{3/2} - 1 = \dfrac{3h + 3h^2 +  h^3}{(1+h)^{3/2} + 1}$ by conjugation, deduce $n\bigl((1 + \tfrac1n)^{3/2} - 1\bigr) \to \tfrac32$, and conclude with Cesàro–Stolz: $$T_n = \sum_{k=1}^{n} \sqrt k \;\sim\; \tfrac23\,  n^{3/2} ,$$ resolving the cliffhanger of question 2.
9. From (G1) alone, derive the logarithm bracketing $$\frac{t}{1 + t} \;\leq\; \ln(1 + t) \;\leq\; t  \qquad (t > -1)$$ *(apply (G1) at $u = \ln(1+t)$ and at $u =  -t/(1+t)$)*.
10. Show that $b_n = \ln n$ is strictly increasing with $\ln n \to +\infty$, and prove with Cesàro–Stolz and question 9 that $$H_n = \sum_{k=1}^{n} \frac 1k \;\sim\; \ln n .$$ (The finer structure $H_n = \ln n + \gamma + o(1)$ is the weekend problem of [Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series).)
11. (From ratios to roots) Let $u_n > 0$ with $\frac{u_{n+1}}{u_n} \to L > 0$ . Using question 9, show $\ln\frac{u_{n+1}}{u_n} \to \ln L$ ; apply Cesàro to conclude $\frac{\ln u_n}{n} \to \ln L$ , then, with (G1), that $u_n^{1/n} \to L$ . Application: compute $\lim\,\binom{2n}{n}^{1/n}$ .

**Part IV — The slow fall of the sine.** Let $u_0 \in \R$ and $u_{n+1} = \sin u_n$.

12. From (G2), show $0 < \sin x < x$ for $0 < x \leq 1$ . Deduce: $u_1 \in \intcc{-1}{1}$ ; if $u_1 = 0$ the sequence is zero from rank $1$ ; and if $u_1 > 0$ (the case $u_1 < 0$ being symmetric, $\sin$ odd), then $(u_n)_{n \geq 1}$ is strictly decreasing, positive, and converges to $0$ *(identify the limit via $\ell =  \sin \ell$, using $\abs{\sin a - \sin b} \leq \abs{a -  b}$, itself a consequence of (G2) and the product-to-sum formula)* .
13. Assume from now on $u_1 \in \intoc{0}{1}$. Show by squeezing, using (G2): $$\frac{\sin u_n}{u_n} \to 1  \qquad\text{and}\qquad  \frac{u_n - \sin u_n}{u_n^{3}} \to \frac16 .$$
14. Prove the factorization $$w_n := \frac{1}{u_{n+1}^{2}} - \frac{1}{u_n^{2}}  = \frac{u_n - \sin u_n}{u_n^{3}} \cdot  \frac{u_n + \sin u_n}{u_n} \cdot  \Bigl(\frac{u_n}{\sin u_n}\Bigr)^{2},$$ and deduce $w_n \to \frac13$.
15. Conclude with [Exercise 11.10](#exo-b1-seq-10) (increments version) that $\frac{1}{n\,u_n^{2}} \to \frac13$, then, by a conjugation argument for the square root, the headline: $$\sqrt n\;u_n \longrightarrow \sqrt 3 ,  \qquad\text{i.e.}\qquad  u_n \sim \sqrt{\frac 3n} .$$
16. Quantify the slowness: show that eventually $\sqrt{2/n}  \leq u_n \leq 2/\sqrt n$ , so that reaching $u_n \leq  10^{-2}$ requires more than $20\,000$ iterations (about $30\,000$ , by the asymptotics). Contrast with [Heron’s method](#ex-b1-seq-heron) ( [Example 11.24](#ex-b1-seq-heron) ) and explain the structural reason: at the fixed point $0$ , the slope of $\sin$ is $1$ (a *neutral* fixed point), while error-halving iterations need a slope of [modulus](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#def-b1-complex-field) $< 1$ .
17. Show that for *every* starting point $u_0 \in \R$ , either $u_n = 0$ from rank $1$ on, or $\abs{u_n} \sim  \sqrt{3/n}$ — the fall is universal, only the sign remembers $u_0$ .

**Part V — The general principle.** The sine is one instance of a machine.

18. Let $u_n > 0$ , $u_n \to 0$ , and $\dfrac{u_n -  u_{n+1}}{u_n^{2}} \to a > 0$ . Prove successively: $\frac{u_{n+1}}{u_n} \to 1$ ; then $\frac{1}{u_{n+1}} -  \frac{1}{u_n} \to a$ ; then $n\,u_n \to \frac1a$ .
19. (Exact model) For $u_{n+1} = \dfrac{u_n}{1 + u_n}$ , $u_0  > 0$ : show $\frac{1}{u_n}$ is arithmetic, solve exactly, and check question 18’s conclusion against the exact formula.
20. For $u_{n+1} = u_n \eu^{-u_n}$ , $u_0 > 0$ : show $u_n \to  0$ , use (G1) to squeeze $\frac{1 - \eu^{-t}}{t}$ between $\frac{1}{1+t}$ and $1$ for $t > 0$ , and conclude $u_n  \sim \frac 1n$ .
21. (Cubic contact, squared telescope) Let $u_n > 0$ , $u_n  \to 0$ , $\dfrac{u_n - u_{n+1}}{u_n^{3}} \to a > 0$ . Adapt question 14’s factorization to show $\frac{1}{u_{n+1}^2} - \frac{1}{u_n^2} \to 2a$ , and conclude $n\,u_n^{2} \to \frac{1}{2a}$ . Check that $a =  \frac16$ recovers Part IV.

**Part VI — Limits of the method, and morals.**

22. Show that the hypothesis $b_n \to +\infty$ cannot be dropped: for $a_n = 2 - 2^{-n}$ and $b_n = 1 - 2^{-n}$ , the increment quotient tends to $1$ while $\frac{a_n}  {b_n} \to 2$ . Point to the exact line of question 5’s proof that breaks.
23. (Stolz twice) Prove $\sum_{k=1}^{n} H_k \sim n \ln n$ *(one application of Cesàro–Stolz, then question 10; bound $(n+1)\ln(n+1) - n\ln n$ using question 9)* .
24. (Geometric means) If $u_n > 0$ and $u_n \to \ell > 0$ , show $(u_1 u_2 \cdots u_n)^{1/n} \to \ell$ ; if $u_n \to  +\infty$ , show $(u_1 \cdots u_n)^{1/n} \to +\infty$ . Deduce $(n!)^{1/n} \to +\infty$ .
25. Synthesis, one sentence each: (i) where exactly did completeness enter this problem; (ii) in what sense is Cesàro–Stolz a discrete l’Hospital rule (its differential twin rests on the mean value theorem of [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative) ); (iii) state the heuristic linking the contact order of $f$ at a neutral fixed point to the decay exponent of $u_{n+1} = f(u_n)$ ; (iv) trace the constant $3$ of $\sqrt{3/n}$ back through the pipeline $\frac16 \to \frac13 \to 3$ .

**Solution of Problem 11.1.**

**1.** $\dfrac{n(n+1)/2}{n^2} = \dfrac{1 + 1/n}{2} \to
\dfrac12$, and $\dfrac{n(n+1)(2n+1)/6}{n^3} = \dfrac{(1 +
1/n)(2 + 1/n)}{6} \to \dfrac13$.

**2.** Upper: each of the $n$ terms is $\leq \sqrt n$, so $T_n \leq n\sqrt n$. Lower: the terms with $k > \frac n2$ number at least $\frac n2$, and each is $\geq \sqrt{n/2}$:

$$
T_n \geq \frac n2 \sqrt{\frac n2} = \frac{n^{3/2}}{2\sqrt 2} .
$$

**3.** For $k \geq N$, since $b_{k+1} - b_k > 0$: $m(b_{k+1} - b_k) \leq a_{k+1} - a_k \leq M(b_{k+1} - b_k)$. Summing for $k = N, \dots, n - 1$, both sides telescope:

$$
m\,(b_n - b_N) \leq a_n - a_N \leq M\,(b_n - b_N),
$$

and dividing by $b_n - b_N > 0$ gives the claim.

**4.** By definition of the limit there is $N$ with $\ell -
\varepsilon \leq \frac{a_{k+1} - a_k}{b_{k+1} - b_k} \leq \ell +
\varepsilon$ for all $k \geq N$; question 3 with $m = \ell -
\varepsilon$, $M = \ell + \varepsilon$ transfers the bracketing to $\frac{a_n - a_N}{b_n - b_N}$.

**5.** Expanding the right side of the identity:

$$
\frac{a_N - \ell b_N}{b_n} + \frac{a_n - a_N}{b_n} -
\ell\,\frac{b_n - b_N}{b_n}
= \frac{a_n - \ell b_n}{b_n} = \frac{a_n}{b_n} - \ell .
$$

By question 4 the second factor of the product is bounded by $\varepsilon$ in absolute value, and $0 < 1 - \frac{b_N}{b_n}
\leq 1$ for large $n$, so

$$
\Bigl|\frac{a_n}{b_n} - \ell\Bigr|
\leq \frac{\abs{a_N - \ell b_N}}{b_n} + \varepsilon
\leq 2\varepsilon
$$

as soon as $b_n \geq \frac{\abs{a_N - \ell b_N}}{\varepsilon}$, which happens eventually since $b_n \to +\infty$. Hence $\frac{a_n}{b_n} \to \ell$: the Cesàro–Stolz theorem.

**6.** Given $M$, choose $N$ with $\frac{a_{k+1} -
a_k}{b_{k+1} - b_k} \geq M$ for $k \geq N$; the lower half of question 3 gives $a_n - a_N \geq M(b_n - b_N)$, so

$$
\frac{a_n}{b_n} \geq \frac{a_N}{b_n} +
M\Bigl(1 - \frac{b_N}{b_n}\Bigr) \longrightarrow M .
$$

Beyond some rank, $\frac{a_n}{b_n} \geq M - 1$; as $M$ was arbitrary, $\frac{a_n}{b_n} \to +\infty$.

**7.** With $b_n = n$ and $a_n = u_1 + \dots + u_n$: the increment quotient is $u_{n+1} \to \ell$, so the Cesàro mean $\frac{a_n}{n}$ tends to $\ell$: part (1) of [Exercise 11.10](#exo-b1-seq-10). With $a_n = u_n$: the increment quotient is $u_{n+1} - u_n$, giving part (3). Converse: $a_n = (-1)^n$, $b_n = n$ has $\frac{a_n}{b_n} \to 0$, yet $a_{n+1} - a_n = \pm
2$ alternates: the increment quotient has no limit.

**8.** Conjugation:

$$
\bigl((1+h)^{3/2} - 1\bigr)\bigl((1+h)^{3/2} + 1\bigr)
= (1+h)^3 - 1 = 3h + 3h^2 + h^3 .
$$

For $h = \frac1n$: $n\bigl((1 + \frac1n)^{3/2} - 1\bigr) =
\frac{3 + 3/n + 1/n^2}{(1 + 1/n)^{3/2} + 1}$, and $1 \leq (1 +
\frac1n)^{3/2} \leq (1 + \frac1n)^2 \to 1$ (squeeze), so the denominator tends to $2$ and the expression to $\frac32$. Now Stolz with $a_n = T_n$, $b_n = n^{3/2}$ (strictly increasing, $\to +\infty$):

$$
\frac{\sqrt{n+1}}{(n+1)^{3/2} - n^{3/2}}
= \frac{\sqrt{n+1}}{\sqrt n} \cdot
\frac{1}{n\bigl((1 + \frac1n)^{3/2} - 1\bigr)}
\longrightarrow 1 \cdot \frac{2}{3},
$$

hence $T_n \sim \frac23\,n^{3/2}$. (The bracketing of question 2 had trapped the constant in $\intcc{0.35}{1}$; Stolz pins it.)

**9.** (G1) at $u = \ln(1+t)$: $1 + t = \eu^{\ln(1+t)} \geq
1 + \ln(1+t)$, so $\ln(1+t) \leq t$. (G1) at $u =
-\frac{t}{1+t}$: $\eu^{-t/(1+t)} \geq 1 - \frac{t}{1+t} =
\frac{1}{1+t} > 0$; taking $\ln$ (increasing): $-\frac{t}{1+t} \geq -\ln(1+t)$, i.e. $\ln(1+t) \geq
\frac{t}{1+t}$.

**10.** $\ln$ is strictly increasing ([Proposition 4.1](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-expln)), and $\ln(2^k) = k\ln 2$ is unbounded, so $\ln n \to +\infty$. Increments: with $t =
\frac1n$ in question 9,

$$
\frac 1{n+1} = \frac{1/n}{1 + 1/n} \leq
\ln\Bigl(1 + \frac1n\Bigr) \leq \frac1n
\quad\Longrightarrow\quad
\frac{n}{n+1} \leq \frac{1/(n+1)}{\ln(1 + 1/n)} \leq 1 ,
$$

so the increment quotient $\frac{H_{n+1} - H_n}{\ln(n+1) - \ln
n}$ tends to $1$; Stolz gives $H_n \sim \ln n$.

**11.** Set $x_n = \frac{u_{n+1}}{u_n} \to L$ and $t_n =
\frac{x_n}{L} - 1 \to 0$. Question 9: $\frac{t_n}{1 + t_n} \leq
\ln(1 + t_n) \leq t_n$, so $\ln x_n - \ln L = \ln(1 + t_n) \to
0$ by squeezing. Cesàro (question 7) applied to $(\ln x_k)$:

$$
\frac1n \sum_{k=0}^{n-1} \ln x_k = \frac{\ln u_n - \ln u_0}{n}
\longrightarrow \ln L ,
$$

so $\frac{\ln u_n}{n} \to \ln L$. With $h_n = \frac{\ln u_n}{n}
- \ln L \to 0$: $u_n^{1/n} = L\,\eu^{h_n}$, and (G1) squeezes $1 + h_n \leq \eu^{h_n} \leq \frac{1}{1 - h_n}$ (for $h_n < 1$), so $\eu^{h_n} \to 1$ and $u_n^{1/n} \to L$. Application: $u_n =
\binom{2n}{n}$ gives

$$
\frac{u_{n+1}}{u_n} = \frac{(2n+1)(2n+2)}{(n+1)^2}
= \frac{2(2n+1)}{n+1} \longrightarrow 4 ,
\qquad\text{so}\qquad
\binom{2n}{n}^{1/n} \to 4 .
$$

**12.** For $0 < x \leq 1$, (G2) gives $\sin x \geq x(1 -
\frac{x^2}{6}) \geq \frac{5x}{6} > 0$ and

$$
x - \sin x \geq \frac{x^3}{6} - \frac{x^5}{120}
= x^3\Bigl(\frac16 - \frac{x^2}{120}\Bigr)
\geq \frac{19}{120}\,x^3 > 0 :
$$

thus $0 < \sin x < x$ on $\intoc{0}{1}$. Always $u_1 = \sin u_0
\in \intcc{-1}{1}$. If $u_1 = 0$, then $u_n = 0$ for $n \geq 1$. If $u_1 \in \intoc{0}{1}$: by induction $0 < u_{n+1} = \sin u_n
< u_n \leq 1$, so $(u_n)_{n\geq1}$ is strictly decreasing and bounded below by $0$: it converges to some $\ell \in
\intco{0}{1}$ ([Theorem 11.9](#thm-b1-seq-monotone)). The product-to-sum formula and (G2) give $\abs{\sin a - \sin b} = 2\abs{\cos
\frac{a+b}{2}}\,\abs{\sin\frac{a-b}{2}} \leq \abs{a - b}$, so $u_{n+1} = \sin u_n \to \sin \ell$: $\ell = \sin\ell$. If $\ell
> 0$ then $\sin\ell < \ell$: impossible. So $u_n \to 0$.

**13.** Dividing (G2) by $u_n > 0$:

$$
1 - \frac{u_n^2}{6} \leq \frac{\sin u_n}{u_n} \leq 1 -
\frac{u_n^2}{6} + \frac{u_n^4}{120} \leq 1 ,
$$

and $u_n \to 0$ squeezes $\frac{\sin u_n}{u_n} \to 1$. Dividing $x - \sin x$ by $x^3$:

$$
\frac16 - \frac{u_n^2}{120} \leq \frac{u_n - \sin u_n}{u_n^3}
\leq \frac16 \longrightarrow \frac16 .
$$

**14.** Since $u_{n+1} = \sin u_n$:

$$
w_n = \frac{u_n^2 - \sin^2 u_n}{u_n^2 \sin^2 u_n}
= \frac{(u_n - \sin u_n)(u_n + \sin u_n)}{u_n^2 \sin^2 u_n}
= \frac{u_n - \sin u_n}{u_n^3}\cdot
\frac{u_n + \sin u_n}{u_n}\cdot
\Bigl(\frac{u_n}{\sin u_n}\Bigr)^{2}
$$

(check the powers of $u_n$: $3 + 1 + (-4)$ against the $u_n^2$ downstairs and $u_n^4$ upstairs). By question 13 the three factors tend to $\frac16$, $2$, $1$: $w_n \to \frac13$.

**15.** $v_n = \frac{1}{u_n^2}$ has increments $v_{n+1} -
v_n = w_n \to \frac13$, so $\frac{v_n}{n} \to \frac13$ by [Exercise 11.10](#exo-b1-seq-10) (3): $n u_n^2 \to 3$. Then

$$
\abs{\sqrt n\,u_n - \sqrt 3}
= \frac{\abs{n u_n^2 - 3}}{\sqrt n\,u_n + \sqrt 3}
\leq \frac{\abs{n u_n^2 - 3}}{\sqrt 3} \longrightarrow 0 :
$$

$\sqrt n\,u_n \to \sqrt 3$, i.e. $u_n \sim \sqrt{3/n}$.

**16.** Since $n u_n^2 \to 3$, eventually $2 \leq n u_n^2
\leq 4$, i.e. $\sqrt{2/n} \leq u_n \leq 2/\sqrt n$. If $u_n
\leq 10^{-2}$ with $n$ in that range, then $2/n \leq 10^{-4}$: $n \geq 20\,000$; and $\sqrt{3/n} = 10^{-2}$ at $n = 30\,000$. [Heron’s method](#ex-b1-seq-heron) squares the error at each step — the digit count doubles — because at its fixed point the relevant slope is $< 1$ in [modulus](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#def-b1-complex-field) (indeed the iteration is contracting). Here $\sin' 0 = \cos 0 = 1$: the fixed point is neutral, no geometric contraction exists, and the decay is governed by the first nonlinear term $-\frac{x^3}{6}$, hence [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def). One step of Heron gains more accuracy than ten thousand steps of the sine.

**17.** For arbitrary $u_0$: $u_1 = \sin u_0 \in
\intcc{-1}{1}$. If $u_1 = 0$ the sequence vanishes from rank $1$. If $u_1 > 0$, Part IV applies verbatim. If $u_1 < 0$, set $v_n = -u_n$: oddness of $\sin$ gives $v_{n+1} = -\sin u_n =
\sin(-u_n) = \sin v_n$ with $v_1 \in \intoc{0}{1}$, so $v_n \sim
\sqrt{3/n}$, i.e. $u_n \sim -\sqrt{3/n}$. In all cases $\abs{u_n} \sim \sqrt{3/n}$ (or the sequence is eventually $0$): the fall is universal, only the sign remembers the start.

**18.** First $\frac{u_{n+1}}{u_n} = 1 - \frac{u_n -
u_{n+1}}{u_n^2}\,u_n \to 1 - a \cdot 0 = 1$. Then

$$
\frac{1}{u_{n+1}} - \frac{1}{u_n}
= \frac{u_n - u_{n+1}}{u_n u_{n+1}}
= \frac{u_n - u_{n+1}}{u_n^2}\cdot\frac{u_n}{u_{n+1}}
\longrightarrow a \cdot 1 = a ,
$$

and [Exercise 11.10](#exo-b1-seq-10) (3) gives $\frac{1}{n u_n} \to a$, i.e. $n u_n \to \frac1a$.

**19.** $v_n = \frac{1}{u_n}$: $v_{n+1} = \frac{1 +
u_n}{u_n} = v_n + 1$, so $v_n = v_0 + n$ and

$$
u_n = \frac{u_0}{1 + n u_0} , \qquad
n u_n = \frac{n u_0}{1 + n u_0} \longrightarrow 1 .
$$

Lemma check: $u_n - u_{n+1} = \frac{u_n^2}{1 + u_n}$, so $\frac{u_n - u_{n+1}}{u_n^2} = \frac{1}{1 + u_n} \to 1 = a$, and question 18 predicts $n u_n \to 1$: exact agreement.

**20.** Positivity by induction ($\eu^{-u} > 0$); decreasing since $\eu^{-u_n} < 1$ for $u_n > 0$; hence $u_n \to
\ell \geq 0$ ([Theorem 11.9](#thm-b1-seq-monotone)). Continuity bridge: with $h_n = \ell - u_n \to 0$, $\eu^{-u_n} = \eu^{-\ell}
\eu^{h_n} \to \eu^{-\ell}$ by the (G1) squeeze $1 + h_n \leq
\eu^{h_n} \leq \frac{1}{1 - h_n}$; so $\ell = \ell\,
\eu^{-\ell}$, and $\ell > 0$ would force $\eu^{-\ell} = 1$, false: $\ell = 0$. For $t > 0$, (G1) gives $\eu^{-t} \geq 1 - t$ and $\eu^{-t} \leq \frac{1}{1 + t}$, so

$$
\frac{1}{1 + t} \leq \frac{1 - \eu^{-t}}{t} \leq 1 .
$$

With $t = u_n$: $\frac{u_n - u_{n+1}}{u_n^2} = \frac{1 -
\eu^{-u_n}}{u_n} \to 1$. Question 18 with $a = 1$: $n u_n \to
1$, so $u_n \sim \frac1n$.

**21.** As in question 18, $\frac{u_{n+1}}{u_n} = 1 -
\frac{u_n - u_{n+1}}{u_n^3}\,u_n^2 \to 1$. Then

$$
\frac{1}{u_{n+1}^2} - \frac{1}{u_n^2}
= \frac{(u_n - u_{n+1})(u_n + u_{n+1})}{u_n^2 u_{n+1}^2}
= \frac{u_n - u_{n+1}}{u_n^3}\cdot
\frac{u_n + u_{n+1}}{u_n}\cdot
\Bigl(\frac{u_n}{u_{n+1}}\Bigr)^2
\longrightarrow a \cdot 2 \cdot 1 = 2a ,
$$

and [Exercise 11.10](#exo-b1-seq-10) (3) gives $\frac{1}{n u_n^2} \to 2a$: $n u_n^2 \to \frac{1}{2a}$. For the sine, $a = \frac16$ (question 13): $n u_n^2 \to 3$, exactly Part IV.

**22.** Increments: $a_{n+1} - a_n = 2^{-n} - 2^{-n-1} =
2^{-n-1} = b_{n+1} - b_n$, so the increment quotient is constantly $1$. Yet $\frac{a_n}{b_n} = \frac{2 - 2^{-n}}{1 -
2^{-n}} \to 2 \neq 1$. The proof of question 5 breaks at the boundary term: $\frac{a_N - \ell b_N}{b_n} \to 0$ needed $b_n
\to +\infty$; here (with $\ell = 1$) $a_N - b_N = 1$ and $b_n
\to 1$, so the term tends to $1$ — precisely the residual gap $2 - 1$.

**23.** Stolz with $A_n = \sum_{k=1}^n H_k$ and $B_n = n\ln
n$: $B_{n+1} - B_n = \ln(n+1) + n\ln(1 + \frac1n) > 0$ and $B_n
\to +\infty$. By question 9, $\frac{n}{n+1} \leq n\ln(1 +
\frac1n) \leq 1$, so $B_{n+1} - B_n = \ln(n+1) + \theta_n$ with $\frac12 \leq \theta_n \leq 1$. Hence

$$
\frac{A_{n+1} - A_n}{B_{n+1} - B_n}
= \frac{H_{n+1}}{\ln(n+1)}\cdot
\frac{1}{1 + \theta_n/\ln(n+1)} \longrightarrow 1 \cdot 1 = 1
$$

(question 10 for the first factor; $\theta_n$ bounded and $\ln(n+1) \to \infty$ for the second). Stolz concludes: $\sum_{k=1}^n H_k \sim n\ln n$.

**24.** If $u_n \to \ell > 0$: as in question 11, $\ln u_n
\to \ln\ell$ (question 9 squeeze on $\ln\frac{u_n}{\ell}$), so the Cesàro means $\frac1n\sum_{k=1}^n \ln u_k \to \ln\ell$, and the exponential bridge of question 11 gives $(u_1\cdots
u_n)^{1/n} = \exp\bigl(\frac1n\sum\ln u_k\bigr) \to \ell$. If $u_n \to +\infty$: for any $M$, eventually $u_n \geq \eu^M$, so $\ln u_n \geq M$: $\ln u_n \to +\infty$; the $+\infty$ Cesàro (question 6, $b_n = n$) gives $\frac1n\sum \ln u_k \to
+\infty$, and (G1) ($\eu^s \geq 1 + s$) sends the geometric mean to $+\infty$. With $u_n = n$: $(n!)^{1/n} \to +\infty$.

**25.** (i) Completeness entered only through the monotone limit theorem, to produce the limits in questions 12 and 20; the Cesàro–Stolz theorem itself is pure $\varepsilon$-management, valid over $\Q$. (ii) Stolz replaces $\lim \frac{a_n}{b_n}$ by $\lim$ of the quotient of increments, exactly as l’Hospital replaces $\lim\frac fg$ by $\lim\frac{f'}{g'}$ — the differential twin rests on the mean value theorem of [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative). (iii) Heuristic: if $f(x) = x - a\,x^{p+1} + o(x^{p+1})$ at the neutral fixed point $0$, then $\frac{1}{u_{n+1}^p} - \frac{1}{u_n^p} \to pa$ and $u_n \sim (pan)^{-1/p}$: contact of order $p + 1$ yields decay $n^{-1/p}$ — the flatter the graph against the diagonal, the slower the fall. (iv) The constant: (G2) supplies the cubic coefficient $\frac16$; the factorization of question 14 doubles it into the telescope increment $\frac13$; Cesàro turns $\frac{1}{u_n^2}$ into $\frac n3$; inverting and taking roots delivers $\sqrt{3/n}$.