Exercises — Sequences — convergence, divergence, boundedness, monotonicity

The figure below plots the first ten terms of three of these sequences on one axis — this is the picture to read the answer off.

How to read the figure: the cyan circles ($a_n=1/n$) march downward and hug the dotted floor at $0$ — decreasing and never leaving $(0,1]$. The amber squares ($b_n=(-1{)}^n$) hop between $+1$ and $-1$ forever — trapped between the two dotted lines but zig-zagging, so no single direction. The white triangles ($d_n=2-1/n$) climb upward toward the dotted ceiling at $2$ without touching it.

(a) $a_n=1,\frac{1}{2},\frac{1}{3},\dots$ — every term lies in $(0,1]$, so bounded. Each term is smaller than the last, so monotone (decreasing). (b) $b_n=-1,1,-1,1,\dots$ — trapped in $[-1,1]$ so bounded, but it jumps up-and-down (amber squares), so not monotone. (c) $c_n=1,2,3,\dots$ — grows past every ceiling, so unbounded (above). Always increasing, so monotone (increasing). (Not plotted — it would run off the top of the frame, which is itself the visual meaning of "unbounded".) (d) $d_n=1,\frac{3}{2},\frac{5}{3},\dots$ climbing toward $2$ but never reaching it (white triangles) — trapped in $[1,2)$ so bounded, and each term exceeds the previous, so monotone (increasing).

(a) $\to 0$ — cyan circles settle onto the floor. (b) diverges — amber squares never settle. (c) diverges (to $\infty$). (d) $\to 2$ — white triangles pack under the ceiling $2$.

WHY divide by $n^2$? The largest power of $n$ dominates for huge $n$; dividing top and bottom by it exposes the pieces that survive versus the pieces that vanish. $\frac{3n^2+1}{n^2-5}=\frac{3+\frac{1}{n^2}}{1-\frac{5}{n^2}}.$ By the algebra of limits, $\frac{1}{n^2}\to 0$ and $\frac{5}{n^2}\to 0$, so the limit is $\frac{3+0}{1-0}=\boxed{3}$.

WHY rationalise? As written it is $\infty -\infty$, an indeterminate shape. Multiplying by the conjugate turns a difference into a fraction we can read. $\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}.$ The denominator grows without bound, so the fraction $\to \boxed{0}$.

WHY squeeze? We cannot track $\cos n$ term-by-term (it wanders), but we can bound it: $\cos n$ never leaves $[-1,1]$. Dividing that band by the positive number $n$ gives $-\frac{1}{n}\le \frac{\cos n}{n}\le \frac{1}{n}.$ The $\epsilon$–$N$ engine behind the Squeeze (spelled out): fix any $\epsilon >0$. The lower bound $-\frac{1}{n}\to 0$ means there is $N_1$ with $n>N_1\Rightarrow -\frac{1}{n}>-\epsilon$; the upper bound $\frac{1}{n}\to 0$ means there is $N_2$ with $n>N_2\Rightarrow \frac{1}{n}<\epsilon$. For $n>N=\max (N_1,N_2)$, chain the sandwich: $-\epsilon <-\frac{1}{n}\le \frac{\cos n}{n}\le \frac{1}{n}<\epsilon ⟹\left|\frac{\cos n}{n}-0\right|<\epsilon .$ That is the $\epsilon$–$N$ condition, so the trapped middle $\to \boxed{0}$. (This is exactly the general Squeeze proof of L5·Q1 applied with $L=0$.)

WHAT we want: given any $\epsilon >0$, produce a cutoff $N$ so that $n>N\Rightarrow |a_n-2|<\epsilon$. First simplify the gap: $|a_n-2|=\left|\frac{2n}{n+1}-2\right|=\left|\frac{2n-2(n+1)}{n+1}\right|=\left|\frac{-2}{n+1}\right|=\frac{2}{n+1}.$ WHY: the gap is a clean formula we can force below $\epsilon$. $\frac{2}{n+1}<\epsilon ⟺n+1>\frac{2}{\epsilon }⟺n>\frac{2}{\epsilon }-1.$ Choose $N=\lceil \frac{2}{\epsilon }\rceil$. Then $n>N\Rightarrow n>\frac{2}{\epsilon }-1\Rightarrow \frac{2}{n+1}<\epsilon$. ✓ Hence $a_n\to 2$.

Method (i) — Squeeze (fully self-contained, uses no external constant). Write the term as a product of $n$ factors: $a_n=\frac{n!}{n^n}=\frac{1\cdot 2\cdot 3\cdots n}{n\cdot n\cdot n\cdots n}=\underset{\text{first factor}}{\underbrace{\frac{1}{n}}}\cdot \underset{\text{each }\le 1}{\underbrace{\frac{2}{n}\cdot \frac{3}{n}\cdots \frac{n}{n}}}.$ Every one of the last $n-1$ factors $\frac{k}{n}$ (for $k=2,\dots ,n$) satisfies $0<\frac{k}{n}\le 1$, so their product is $\le 1$. Hence $0<a_n\le \frac{1}{n}.$ Both outer bounds ($0$ and $\frac{1}{n}$) tend to $0$, so by the Squeeze Theorem $a_n\to \boxed{0}$. This method is complete on its own and introduces no new constant.

Method (ii) — ratio, with an honest geometric-decay proof. With factorials and powers, comparing consecutive terms is easy: $\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1{)}^{n+1}}\cdot \frac{n^n}{n!}=\frac{(n+1)n^n}{(n+1{)}^{n+1}}=\frac{n^n}{(n+1{)}^n}={\left(\frac{n}{n+1}\right)}^n.$ We must now turn "the ratio is small" into an actual proof of decay. The cleanest self-contained bound: for every $n\ge 1$, $\frac{a_{n+1}}{a_n}={\left(\frac{n}{n+1}\right)}^n\le \frac{n}{n+1}<1,$ because raising a number in $(0,1)$ to the power $n\ge 1$ only makes it $\le$ itself (multiplying by more factors each $<1$). But we can pin a fixed ceiling well below $1$: for all $n\ge 1$, $\frac{n}{n+1}\le \frac{1}{2}$ is false (it exceeds $\frac{1}{2}$ from $n=2$ on), so instead evaluate at the very first index to get a uniform bound valid for all $n\ge 1$: $\frac{a_{n+1}}{a_n}={\left(\frac{n}{n+1}\right)}^n\le {\left(\frac{n}{n+1}\right)}^1\le \frac{1}{2}\text{is invalid for large }n;$ so we take a genuinely uniform ceiling: choose the constant $q$ from the second step onward. For $n\ge 2$, $\frac{a_{n+1}}{a_n}={\left(\frac{n}{n+1}\right)}^n\le {\left(\frac{n}{n+1}\right)}^2\le {\left(\frac{2}{3}\right)}^2=\frac{4}{9}=:q<1.$ Here we used two monotone facts: raising the base $<1$ to a larger power ($n\ge 2$) makes it smaller, and $\frac{n}{n+1}$ increases in $n$ so at $n=2$ it is largest among... no — $\frac{n}{n+1}$ increases toward $1$, so we cannot bound it above by its value at $n=2$. To get a correct uniform ceiling we keep the exponent fixed at $2$ and bound only the drop from one factor: since ${\left(\frac{n}{n+1}\right)}^n\le {\left(\frac{n}{n+1}\right)}^2$ and this last quantity is $<1$ for every $n\ge 2$ but creeps up to $1$, no single $q<1$ works uniformly from a fixed base. Therefore we localise the ceiling instead of globalising it, which is exactly what a limit provides — and we now supply that limit without invoking $e$: ${\left(\frac{n}{n+1}\right)}^n=\frac{1}{{\left(1+\frac{1}{n}\right)}^n}.$ The denominator ${\left(1+\frac{1}{n}\right)}^n$ is an increasing sequence (this is the L4-style monotonicity already used in the parent note) and is bounded above by $3$ (parent note, binomial bound). Hence ${\left(1+\frac{1}{n}\right)}^n\ge {\left(1+\frac{1}{1}\right)}^1=2$ for all $n\ge 1$, so $\frac{a_{n+1}}{a_n}=\frac{1}{{\left(1+\frac{1}{n}\right)}^n}\le \frac{1}{2}=:q<1\text{for all }n\ge 1.$ That is the honest uniform ceiling. Now conclude geometric decay: for every $n\ge 1$, $a_{n+1}\le q{a}_n⟹a_n\le q{a}_{n-1}\le q^2{a}_{n-2}\le \cdots \le q^{n-1}{a}_1={\left(\frac{1}{2}\right)}^{n-1}\cdot \frac{1!}{1^1}={\left(\frac{1}{2}\right)}^{n-1}.$ Since $0<q<1$, $q^{n-1}\to 0$, so $0<a_n\le q^{n-1}\to 0$ and by Squeeze $a_n\to 0$ again. No undefined constant is used anywhere: the ceiling $q=\frac{1}{2}$ came only from ${\left(1+\frac{1}{n}\right)}^n\ge 2$, a bare inequality about the first term of an increasing sequence. (For readers curious where this sequence's limit lives, see The number e — but the proof above never needs its value.)

Follow the parent's recipe with $L=2,\epsilon =1$. From L3·Q1, $|a_n-2|=\frac{2}{n+1}\le 1$ for all $n\ge 1$, so every term sits in $[1,3)$: the tail is bounded by $|L|+1=3$. The (single) head term is $a_1=1$. Take $K=\max (3,|a_1|)=\boxed{3}$: $|a_n|\le 3$ for all $n$.

We combine boundedness + monotonicity ⟹ MCT, then solve a fixed point. The staircase figure shows the plan; walk through it after reading Steps 1–4.

Step 0 — positivity (needed later). Every term is positive: $a_1=1>0$, and if $a_n>0$ then $a_{n+1}=\sqrt{2+a_n}>0$ (a square root of a positive number). By induction $a_n>0$ for all $n$. WHY bother: Step 2 will compare $a_{n+1}$ and $a_n$ through their squares, and squaring only preserves order for nonnegative numbers — so we must know both are positive first. Step 1 — bounded above by 2 (induction). $a_1=1\le 2$. If $a_n\le 2$ then $a_{n+1}=\sqrt{2+a_n}\le \sqrt{2+2}=2$. So $a_n\le 2$ for all $n$. WHY 2? It is the suspected limit; testing the ceiling you expect is the standard move. Step 2 — increasing. Compute the difference of squares: $a_{n+1}^2-a_n^2=(2+a_n)-a_n^2=-(a_n^2-a_n-2)=-(a_n-2)(a_n+1).$ Since $a_n\le 2$ we have $(a_n-2)\le 0$, and since $a_n>0$ (Step 0) we have $(a_n+1)>0$; their product is $\le 0$, so $-(a_n-2)(a_n+1)\ge 0$. Hence $a_{n+1}^2\ge a_n^2$. Because both $a_{n+1}>0$ and $a_n>0$, squaring preserved the order, so $a_{n+1}\ge a_n$. WHY square? The recursion has a square root; squaring removes it so the sign is readable. Step 3 — MCT. Increasing + bounded above $\Rightarrow$ converges to some $L$. Step 4 — solve the fixed point. Both $a_n\to L$ and $a_{n+1}\to L$, and $a_{n+1}=\sqrt{2+a_n}$ gives $L=\sqrt{2+L}$, i.e. $L^2-L-2=0\Rightarrow (L-2)(L+1)=0$. Since terms are positive, $L=\boxed{2}$.

Reading the staircase figure: start at the amber dot on the horizontal axis at $a_1=1$. Move vertically up to the cyan curve $y=\sqrt{2+x}$ — that height is $a_2$. Move horizontally across to the white diagonal $y=x$ to copy $a_2$ back onto the axis, then vertically again to the curve for $a_3$, and so on. The amber staircase climbs rightward and upward, its steps shrinking as it homes in on the white dot where curve meets diagonal — the fixed point $L=2$.

Boundedness: $\left|\frac{(-1{)}^{n}n}{n+1}\right|=\frac{n}{n+1}<1$, so $(b_n)$ is bounded. But that alone does not grant convergence. Split by parity. For even $n$: $b_n=\frac{n}{n+1}\to 1$. For odd $n$: $b_n=\frac{-n}{n+1}\to -1$. The even-indexed subsequence approaches $1$, the odd-indexed one approaches $-1$. A convergent sequence forces every subsequence to the same limit; here two subsequences disagree, so $(b_n)$ diverges. Bolzano–Weierstrass angle: the sequence is bounded, so B–W guarantees some convergent subsequence exists — indeed we found two (limits $1$ and $-1$). Boundedness buys convergent subsequences, not convergence of the whole.

Goal: given $\epsilon >0$, find $N$ with $n>N\Rightarrow |a_n-L|<\epsilon$. Since $b_n\to L$: $\exists N_1$ with $n>N_1\Rightarrow L-\epsilon <b_n$. Since $c_n\to L$: $\exists N_2$ with $n>N_2\Rightarrow c_n<L+\epsilon$. WHY two cutoffs then a max? Each limit gives its own wall; taking $N=\max (N_1,N_2)$ makes both promises hold at once. For $n>N=\max (N_1,N_2)$, chain the sandwich: $L-\epsilon <b_n\le a_n\le c_n<L+\epsilon ⟹|a_n-L|<\epsilon .$ This is exactly the $\epsilon$–$N$ condition, so $a_n\to L$. $■$

Strategy (proof by contradiction): suppose $A\ne A^{\prime }$. Let $\epsilon =\frac{|A-A^{\prime }|}{2}>0$ — half the gap between the two candidate limits. WHY half? So the two bands $(A-\epsilon ,A+\epsilon )$ and $(A^{\prime }-\epsilon ,A^{\prime }+\epsilon )$ cannot overlap. Convergence to $A$: $\exists N_1$, $n>N_1\Rightarrow |a_n-A|<\epsilon$. Convergence to $A^{\prime }$: $\exists N_2$, $n>N_2\Rightarrow |a_n-A^{\prime }|<\epsilon$. Combine at one index. Pick any $n>N=\max (N_1,N_2)$; then both inequalities hold at that single $n$. Now use the triangle inequality $|x+y|\le |x|+|y|$ with $x=A-a_n$ and $y=a_n-A^{\prime }$ (note $x+y=A-A^{\prime }$): $|A-A^{\prime }|=|(A-a_n)+(a_n-A^{\prime })|\le |A-a_n|+|a_n-A^{\prime }|<\epsilon +\epsilon =2\epsilon =|A-A^{\prime }|.$ This says $|A-A^{\prime }|<|A-A^{\prime }|$, which is impossible (no number is strictly less than itself). The impossibility came only from assuming $A\ne A^{\prime }$ (that was what made $\epsilon >0$). Therefore $A=A^{\prime }$: a sequence has at most one limit. $■$

We use only elementary algebra plus MCT — no external inequality is assumed; the one bound we need is proved from scratch below.

Positivity. $x_1=2>0$; if $x_n>0$ then $x_{n+1}=\frac{1}{2}(x_n+\frac{2}{x_n})$ is an average of positive numbers, so $x_{n+1}>0$. By induction all terms are positive.

Step 1 — bounded below by $\sqrt{2}$ (proved directly, not by quoting AM–GM). For any $x_n>0$, $x_{n+1}^2-2={\left(\frac{1}{2}\left(x_n+\frac{2}{x_n}\right)\right)}^2-2=\frac{x_n^2}{4}+1+\frac{1}{x_n^2}-2=\frac{x_n^2}{4}-1+\frac{1}{x_n^2}={\left(\frac{x_n}{2}-\frac{1}{x_n}\right)}^2\ge 0.$ The key move is recognising the expression as a perfect square, which is $\ge 0$ for free. So $x_{n+1}^2\ge 2$, and since $x_{n+1}>0$, $x_{n+1}\ge \sqrt{2}$ for every $n\ge 1$. Step 2 — decreasing (from $n\ge 2$). Using $x_n^2\ge 2$ (Step 1): $x_n-x_{n+1}=x_n-\frac{1}{2}\left(x_n+\frac{2}{x_n}\right)=\frac{x_n^2-2}{2x_n}\ge 0,$ because the numerator $x_n^2-2\ge 0$ and the denominator $2x_n>0$. So the sequence decreases. Step 3 — MCT. Decreasing + bounded below $\Rightarrow$ converges to $L=\inf x_n$; existence of this infimum is precisely the Completeness of $\mathbb{R}$ (greatest-lower-bound property). Solve $L=\frac{1}{2}(L+\frac{2}{L})\Rightarrow L^2=2\Rightarrow L=\boxed{\sqrt{2}}$ (positive root, since all terms $>0$). Cauchy angle: in $\mathbb{R}$, "convergent" and "Cauchy" mean the same thing (again by completeness). This sequence is Cauchy — its terms bunch arbitrarily close — yet if we lived in $\mathbb{Q}$ alone, the target $\sqrt{2}$ would be missing, and the sequence would be a Cauchy sequence with no limit. Completeness is exactly what plugs that hole.

To prove a recursion converges, in order? ::: (1) positivity if needed, (2) bounded, (3) monotone $\Rightarrow$ MCT, (4) solve $L=f(L)$. Bounded alone gives you what (not convergence)? ::: A convergent subsequence (Bolzano–Weierstrass). In $\epsilon$–$N$ proofs $N$ may depend on? ::: $\epsilon$ (and $L$) only — never on the running index $n$. Two subsequences with different limits imply? ::: The full sequence diverges. Why does Heron's sequence need completeness? ::: Its limit $\sqrt{2}\notin \mathbb{Q}$; only complete $\mathbb{R}$ supplies it.