Lecture 21

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


The sequence a1=0,a2=1/2,a3=2/3,a4=3/4,a5=4/5,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 an=n1na_n = \frac{n-1}{n}, converges to 11.


[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 ϵ>0\epsilon>0, there is some NN, such that for all n>Nn>N we have n1n1<ϵ.\left| \frac{n-1}{n} - 1 \right| < \epsilon.

A natural thing to do is to simplify that: n1n1=(n1)nn=1n=1n.\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 ϵ>0\epsilon>0 there is some NN, such that for all n>Nn>N we have 1n<ϵ.\frac{1}{n} < \epsilon. But that’s the same as having n>1ϵ.n > \frac{1}{\epsilon}.

So if we take NN to be 1ϵ\left\lceil\frac{1}{\epsilon}\right\rceil, the smallest integer greater than 1/ϵ1/\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 (neat version)] We must show that, for every ϵ>0\epsilon>0, there is some NN such that for all n>Nn>N we have n1n1<ϵ.\left| \frac{n-1}{n} - 1 \right| < \epsilon. Let such an ϵ\epsilon be given.

Define NN to be 1ϵ\left\lceil\frac{1}{\epsilon}\right\rceil, which is the smallest integer greater than 1/ϵ1/\epsilon.

Then, if n>Nn>N, we have n1n1=(n1)nn=1n=1n<1N<11/ϵ=ϵ,\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:


The sequence defined by an=3n+13n+1a_n = \frac{3^{n+1}}{3^n+1} converges to 33.


[Rough version] We need to show that, for all ϵ>0\epsilon>0, there exists some NN such that, for all n>Nn>N, we have an3<ϵ.\left|a_n - 3\right| < \epsilon.

We can simplify the left-hand-side considerably: an3=3n+13n+13=3n+13n+13(3n+1)3n+1=3n+13(3n+1)3n+1=3n+13n+133n+1=33n+1=33n+1.\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 ϵ>0\epsilon>0, there exists some NN such that, for all n>Nn>N, we have 33n+1<ϵ.\frac{3}{3^n+1} < \epsilon.

We can rearrange this to 3ϵ<3n+1,and then3ϵ1<3n,and thenlog3(3/ϵ1)<n.\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=log3(3/ϵ1)N= \lceil \log_3(3/\epsilon-1) \rceil, the smallest integer greater than log3(3/ϵ1)\log_3(3/\epsilon-1). this will work.


A neater version

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


[Neat version] We need to show that, for all ϵ>0\epsilon>0, there exists some NN such that, for all n>Nn>N, we have an3<ϵ.\left|a_n - 3\right| < \epsilon. Suppose given some ϵ\epsilon; we’ll show that N=log3(3/ϵ1)N = \lceil \log_3(3/\epsilon-1) \rceil works. Indeed, we have an3=3n+13n+13=3n+13(3n+1)3n+1=3n+13n+133n+1=33n+1=33n+1<33N+1=33log3(3/ϵ1)+133log3(3/ϵ1)+1=33/ϵ1+1=33/ϵ=ϵ,\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:


[Sandwich Lemma] Suppose we have three sequences: a0,a1,a2,a_0, a_1, a_2, \ldots, and b0,b1,b2,b_0, b_1, b_2, \ldots and c0,c1,c2,c_0, c_1, c_2, \ldots, such that:

  1. the sequences (ai)iN(a_i)_{i\in\mathbb{N}} and (ci)iN(c_i)_{i\in\mathbb{N}} both converge to the same number xx; and

  2. for all ii we have aibicia_i \leq b_i \leq c_i

Then the sequence (bi)iN(b_i)_{i\in\mathbb{N}} also converges to xx.


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

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

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

Given that, I claim we can take NN to be max(M,P)\max(M,P), the larger of MM and PP. Then we have bnϵcnϵ<x<an+ϵbn+ϵ,b_n-\epsilon \leq c_n-\epsilon < x < a_n+\epsilon \leq b_n+\epsilon, as required. Here the outer two inequalities are because aibicia_i\leq b_i\leq c_i for all ii, and the inner two are obtained from the convergence of (ai)iN(a_i)_{i\in\mathbb{N}} and (ci)iN(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 bn=3+sin(n7nn)nb_n = 3 + \frac{\sin(n^{7n}-\sqrt{n})}{n} converges to 33.

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

However, we can proceed by sandwiching it between an=31na_n = 3-\frac{1}{n} and cnc_n = 3+1n3+\frac{1}{n} (since all values of sin\sin are always between 1-1 and 11). Showing that those two sequences both converge to 33 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:


Let a0,a1,a_0,a_1,\ldots be a sequence that converges to xx, and let b0,b1,b_0,b_1,\ldots be a sequence that converges to yy. Then the sequence a0+b0,a1+b1,a_0+b_0,\quad a_1+b_1,\quad\ldots converges to x+yx+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.


Adding sequences


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

Since the sequence (ai)iN(a_i)_{i\in\mathbb{N}} converges to xx, there is some RR such that for all n>Rn>R we have anx<ϵ2,\left|a_n-x\right| < \frac{\epsilon}{2}, and since (bi)iN(b_i)_{i\in\mathbb{N}} converges to yy, there is some SS such that for all n>Sn>S we have bny<ϵ2.\left|b_n-y\right| < \frac{\epsilon}{2}.

That means that, if we take N=max(R,S)N=\max(R,S) then for all n>Nn>N we have n>Rn>R and n>Sn>S, and so (an+bn)(x+y)=(anx)+(bny)anx+bny(by the triangle inequality)<ϵ2+ϵ2(since n>R and n>S)=ϵ,\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 a0b0,a1b1,a2b2,a_0b_0,\quad a_1b_1,\quad a_2b_2,\quad\ldots converges to xyxy, and that (if yy and all the terms bib_i are nonzero), that a0/b0,a1/b1,a2/b2,a_0/b_0,\quad a_1/b_1,\quad a_2/b_2,\quad\ldots converges to x/yx/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 xx if, no matter what you mean by “close”, the sequence eventually gets close to xx and stays close to xx forever. In symbols, that was ϵ>0,NNs.t.n>N,anx<ϵ.\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 a0,a1,a2,a_0, a_1, a_2, \ldots is said to be Cauchy if ϵ>0,NNs.t.m,n>N,aman<ϵ.\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.