Lecture 21

Let’s now try proving that some sequence or other does converge, as we’re not well practiced at that yet:

Proposition

The sequence $a_1=0,\quad a_2=1/2, \quad a_3=2/3, \quad a_4=3/4, \quad a_5=4/5,\quad\ldots$ where $a_n = \frac{n-1}{n}$, converges to $1$.

Proof

[Proof (rough version)]

The definition of convergence is complicated, so it may be helpful to start by reminding us what we’re aiming for. So we’ll start by working from the wrong end.

We need to show that, for every $\epsilon>0$, there is some $N$, such that for all $n>N$ we have $\left| \frac{n-1}{n} - 1 \right| < \epsilon.$

A natural thing to do is to simplify that: $\left|\frac{n-1}{n} - 1 \right| {= \left| \frac{(n-1) - n}{n}\right|} {= \left|\frac{-1}{n}\right|} {= \frac{1}{n}}.$

So, as we can see, what we’re aiming for is that, for every $\epsilon>0$ there is some $N$, such that for all $n>N$ we have $\frac{1}{n} < \epsilon.$ But that’s the same as having $n > \frac{1}{\epsilon}.$

So if we take $N$ to be $\left\lceil\frac{1}{\epsilon}\right\rceil$, the smallest integer greater than $1/\epsilon$, that works.

That proof is sort-of-okay, but it’s backwards. It was helpful to write it, but hard to check that it’s logically valid. I’ll now rewrite it forwards.

Proof

[Proof (neat version)] We must show that, for every $\epsilon>0$, there is some $N$ such that for all $n>N$ we have $\left| \frac{n-1}{n} - 1 \right| < \epsilon.$ Let such an $\epsilon$ be given.

Define $N$ to be $\left\lceil\frac{1}{\epsilon}\right\rceil$, which is the smallest integer greater than $1/\epsilon$.

Then, if $n>N$, we have $\left| \frac{n-1}{n} - 1 \right| {}= \left| \frac{(n-1) - n}{n}\right| {}= \left|\frac{-1}{n}\right| {}= \frac{1}{n} {}< \frac{1}{N} {}< {}\frac{1}{1/\epsilon} {}= \epsilon,$ exactly as required.

That second version is obviously correct, and all the reasoning goes in the right direction. But analysis proofs often have the property that the most persuasive proof will seem a bit mysterious. It’s best to do the rough work and then rewrite it neatly.

Let’s do another example:

Proposition

The sequence defined by $a_n = \frac{3^{n+1}}{3^n+1}$ converges to $3$.

Proof

[Rough version] We need to show that, for all $\epsilon>0$, there exists some $N$ such that, for all $n>N$, we have $\left|a_n - 3\right| < \epsilon.$

We can simplify the left-hand-side considerably: $\begin{gathered} \left|a_n - 3\right| {= \left|\frac{3^{n+1}}{3^n+1} - 3\right|} {= \left|\frac{3^{n+1}}{3^n+1} - \frac{3(3^n+1)}{3^n+1}\right|}\\ {= \left|\frac{3^{n+1}-3(3^n+1)}{3^n+1}\right|} {= \left|\frac{3^{n+1}-3^{n+1}-3}{3^n+1}\right|} {= \left|\frac{-3}{3^n+1}\right|} {= \frac{3}{3^n+1}.}\end{gathered}$

So we need to show that, for all $\epsilon>0$, there exists some $N$ such that, for all $n>N$, we have $\frac{3}{3^n+1} < \epsilon.$

We can rearrange this to $\frac{3}{\epsilon} < 3^n+1, {\quad\text{and then}\quad \frac{3}{\epsilon} - 1 < 3^n,} {\quad\text{and then}\quad \log_3(3/\epsilon - 1) < n.}$

Hence if we take $N= \lceil \log_3(3/\epsilon-1) \rceil$, the smallest integer greater than $\log_3(3/\epsilon-1)$. this will work.

[t]

A neater version

As before, here’s a shorter, more persuasive but less insightful version of the same material:

Proof

[Neat version] We need to show that, for all $\epsilon>0$, there exists some $N$ such that, for all $n>N$, we have $\left|a_n - 3\right| < \epsilon.$ Suppose given some $\epsilon$; we’ll show that $N = \lceil \log_3(3/\epsilon-1) \rceil$ works. Indeed, we have $\begin{gathered} \left|a_n - 3\right| {= \left|\frac{3^{n+1}}{3^n+1} - 3\right|} {= \left|\frac{3^{n+1}-3(3^n+1)}{3^n+1}\right|}\\ {= \left|\frac{3^{n+1}-3^{n+1}-3}{3^n+1}\right|} {= \left|\frac{-3}{3^n+1}\right|} {= \frac{3}{3^n+1}} {< \frac{3}{3^N+1}}\\ {= \frac{3}{3^{\lceil\log_3(3/\epsilon-1)\rceil}+1}} {\leq \frac{3}{3^{\log_3(3/\epsilon-1)}+1}} {= \frac{3}{3/\epsilon-1+1}} {= \frac{3}{3/\epsilon}} {= \epsilon,}\end{gathered}$ exactly as required.

Some more systematic methods

The examples above seem like a lot of work. In fact, they are a lot of work: the definition of convergence is genuinely complicated, so it’s no surprise that it takes time when you have to use it.

However, real analysts know many situations that crop up repeatedly, and know many useful facts for saving them time in those situations.

One useful one is the following:

Theorem

[Sandwich Lemma] Suppose we have three sequences: $a_0, a_1, a_2, \ldots$, and $b_0, b_1, b_2, \ldots$ and $c_0, c_1, c_2, \ldots$, such that:

1. the sequences $(a_i)_{i\in\mathbb{N}}$ and $(c_i)_{i\in\mathbb{N}}$ both converge to the same number $x$; and

2. for all $i$ we have $a_i \leq b_i \leq c_i$

Then the sequence $(b_i)_{i\in\mathbb{N}}$ also converges to $x$.

Proof

We must show that, for all $\epsilon$, there is an $N$ such that, for all $n>N$ $\left|b_n-x\right|<\epsilon,$ or, to put it differently, for all $n>N$ we have $b_n-\epsilon < x < b_n+\epsilon.$

Suppose given such an $\epsilon$. Since $(a_i)_{i\in\mathbb{N}}$ converges to $x$, there is an $M$ such that, for all $n>M$, we have $\left|a_n-x\right|<\epsilon,$ or in other words $a_n-\epsilon < x < a_n+\epsilon.$

Since $(c_i)_{i\in\mathbb{N}}$ converges to $x$, there is also an $P$ such that, for all $n>P$, we have $\left|c_n-x\right|<\epsilon,$ or in other words $c_n-\epsilon < x < c_n+\epsilon.$

Given that, I claim we can take $N$ to be $\max(M,P)$, the larger of $M$ and $P$. Then we have $b_n-\epsilon \leq c_n-\epsilon < x < a_n+\epsilon \leq b_n+\epsilon,$ as required. Here the outer two inequalities are because $a_i\leq b_i\leq c_i$ for all $i$, and the inner two are obtained from the convergence of $(a_i)_{i\in\mathbb{N}}$ and $(c_i)_{i\in\mathbb{N}}$.

The usefulness of this result is that we can ignore complicated features of sequences by “sandwiching” them between simpler things.

For example, using this we can show that the sequence $b_n = 3 + \frac{\sin(n^{7n}-\sqrt{n})}{n}$ converges to $3$.

The mess inside the $\sin$ brackets is extremely unpleasant, and we’d love to not have to work with it.

However, we can proceed by sandwiching it between $a_n = 3-\frac{1}{n}$ and $c_n$ = $3+\frac{1}{n}$ (since all values of $\sin$ are always between $-1$ and $1$). Showing that those two sequences both converge to $3$ is no harder than the examples we’ve done already.

Another sensible question to ask is to do with combining sequences. Here we’ll talk about addition:

Theorem

Let $a_0,a_1,\ldots$ be a sequence that converges to $x$, and let $b_0,b_1,\ldots$ be a sequence that converges to $y$. Then the sequence $a_0+b_0,\quad a_1+b_1,\quad\ldots$ converges to $x+y$.

I’ll give just a neat version of the proof. The idea is that in order for a sum to be close, we can tolerate each summand being up to half the permitted distance away.

[t]

Adding sequences

Proof

We must show that for all $\epsilon>0$, there is some $N$ such that for all $n>N$ we have $\left|(a_n+b_n)-(x+y)\right| < \epsilon.$ Suppose we are given $\epsilon$.

Since the sequence $(a_i)_{i\in\mathbb{N}}$ converges to $x$, there is some $R$ such that for all $n>R$ we have $\left|a_n-x\right| < \frac{\epsilon}{2},$ and since $(b_i)_{i\in\mathbb{N}}$ converges to $y$, there is some $S$ such that for all $n>S$ we have $\left|b_n-y\right| < \frac{\epsilon}{2}.$

That means that, if we take $N=\max(R,S)$ then for all $n>N$ we have $n>R$ and $n>S$, and so $\begin{aligned} \left|(a_n+b_n)-(x+y)\right| & {= \left|(a_n-x)+(b_n-y)\right|}\\ & {\leq \left|a_n-x\right|+\left|b_n-y\right|}\\ & {\qquad\text{(by the triangle inequality)}}\\ & {<\frac{\epsilon}{2}+\frac{\epsilon}{2}}\\ & {\qquad\text{(since $n>R$ and $n>S$)}}\\ & {=\epsilon,}\end{aligned}$ which is what we needed.

A very similar argument will do subtraction.

It’s also true that the sequence $a_0b_0,\quad a_1b_1,\quad a_2b_2,\quad\ldots$ converges to $xy$, and that (if $y$ and all the terms $b_i$ are nonzero), that $a_0/b_0,\quad a_1/b_1,\quad a_2/b_2,\quad\ldots$ converges to $x/y$. These are slightly (but not much) harder.

There are many, many more facts about convergent sequences, but I’ll leave it there. Ultimately the aim should not be to memorise facts, but to have enough tools that you can prove them yourself, as needed.

Cauchy sequences

We have said what it means for a sequence to converge to a number.

We’re now going to explore a concept which says that a sequence “ought to converge to something (but we’re not sure what)”.

Recall that a sequence converges to $x$ if, no matter what you mean by “close”, the sequence eventually gets close to $x$ and stays close to $x$ forever. In symbols, that was $\forall \epsilon>0,\quad \exists N\in\mathbb{N}\quad\mathrm{s.t.}\quad\forall n>N,\quad\left|a_n-x\right|<\epsilon.$

If a sequence ought to converge to something, then its terms ought to “settle down” somehow. That means they should at least get close at least to each other.

This leads us to the following definition:
Definition: A sequence $a_0, a_1, a_2, \ldots$ is said to be Cauchy if $\forall \epsilon>0,\quad \exists N\in\mathbb{N}\quad\mathrm{s.t.}\quad\forall m,n>N,\quad\left|a_m-a_n\right|<\epsilon.$

In vaguer terms, a sequence is Cauchy if no matter what we mean by close, there is some point beyond which all the terms of the sequence are close to each other.