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,… where an=nn−1, converges to 1.
[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, there is some N, such that for all n>N we have ∣∣∣∣nn−1−1∣∣∣∣<ϵ.
A natural thing to do is to simplify that:∣∣∣∣nn−1−1∣∣∣∣=∣∣∣∣n(n−1)−n∣∣∣∣=∣∣∣∣n−1∣∣∣∣=n1.
So, as we can see, what we’re aiming for is that, for every ϵ>0 there is some N, such that for all n>N we have n1<ϵ.But that’s the same as having n>ϵ1.
So if we take N to be ⌈ϵ1⌉, the smallest integer greater than 1/ϵ, 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, there is some N such that for all n>N we have ∣∣∣∣nn−1−1∣∣∣∣<ϵ. Let such an ϵ be given.
Define N to be ⌈ϵ1⌉, which is the smallest integer greater than 1/ϵ.
Then, if n>N, we have ∣∣∣∣nn−1−1∣∣∣∣=∣∣∣∣n(n−1)−n∣∣∣∣=∣∣∣∣n−1∣∣∣∣=n1<N1<1/ϵ1=ϵ, 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+1 converges to 3.
[Rough version] We need to show that, for all ϵ>0, there exists some N such that, for all n>N, we have ∣an−3∣<ϵ.
We can simplify the left-hand-side considerably: ∣an−3∣=∣∣∣∣3n+13n+1−3∣∣∣∣=∣∣∣∣3n+13n+1−3n+13(3n+1)∣∣∣∣=∣∣∣∣3n+13n+1−3(3n+1)∣∣∣∣=∣∣∣∣3n+13n+1−3n+1−3∣∣∣∣=∣∣∣∣3n+1−3∣∣∣∣=3n+13.
So we need to show that, for all ϵ>0, there exists some N such that, for all n>N, we have 3n+13<ϵ.
We can rearrange this to ϵ3<3n+1,and thenϵ3−1<3n,and thenlog3(3/ϵ−1)<n.
Hence if we take N=⌈log3(3/ϵ−1)⌉, the smallest integer greater than log3(3/ϵ−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, there exists some N such that, for all n>N, we have ∣an−3∣<ϵ. Suppose given some ϵ; we’ll show that N=⌈log3(3/ϵ−1)⌉ works. Indeed, we have ∣an−3∣=∣∣∣∣3n+13n+1−3∣∣∣∣=∣∣∣∣3n+13n+1−3(3n+1)∣∣∣∣=∣∣∣∣3n+13n+1−3n+1−3∣∣∣∣=∣∣∣∣3n+1−3∣∣∣∣=3n+13<3N+13=3⌈log3(3/ϵ−1)⌉+13≤3log3(3/ϵ−1)+13=3/ϵ−1+13=3/ϵ3=ϵ,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,…, and b0,b1,b2,… and c0,c1,c2,…, such that:
the sequences (ai)i∈N and (ci)i∈N both converge to the same number x; and
for all i we have ai≤bi≤ci
Then the sequence (bi)i∈N also converges to x.
We must show that, for all ϵ, there is an N such that, for all n>N∣bn−x∣<ϵ, or, to put it differently, for all n>N we have bn−ϵ<x<bn+ϵ.
Suppose given such an ϵ. Since (ai)i∈N converges to x, there is an M such that, for all n>M, we have ∣an−x∣<ϵ, or in other words an−ϵ<x<an+ϵ.
Since (ci)i∈N converges to x, there is also an P such that, for all n>P, we have ∣cn−x∣<ϵ, or in other words cn−ϵ<x<cn+ϵ.
Given that, I claim we can take N to be max(M,P), the larger of M and P.Then we have bn−ϵ≤cn−ϵ<x<an+ϵ≤bn+ϵ, as required. Here the outer two inequalities are because ai≤bi≤ci for all i, and the inner two are obtained from the convergence of (ai)i∈N and (ci)i∈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+nsin(n7n−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 an=3−n1 and cn = 3+n1 (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:
Let a0,a1,… be a sequence that converges to x, and let b0,b1,… be a sequence that converges to y. Then the sequence a0+b0,a1+b1,… 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.
We must show that for all ϵ>0, there is some N such that for all n>N we have ∣(an+bn)−(x+y)∣<ϵ.Suppose we are given ϵ.
Since the sequence (ai)i∈N converges to x, there is some R such that for all n>R we have ∣an−x∣<2ϵ, and since (bi)i∈N converges to y, there is some S such that for all n>S we have ∣bn−y∣<2ϵ.
That means that, if we take N=max(R,S) then for all n>N we have n>R and n>S, and so ∣(an+bn)−(x+y)∣=∣(an−x)+(bn−y)∣≤∣an−x∣+∣bn−y∣(by the triangle inequality)<2ϵ+2ϵ(since n>R and n>S)=ϵ,which is what we needed.
A very similar argument will do subtraction.
It’s also true that the sequence a0b0,a1b1,a2b2,… converges to xy, and that (if y and all the terms bi are nonzero), that a0/b0,a1/b1,a2/b2,… 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.
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 ∀ϵ>0,∃N∈Ns.t.∀n>N,∣an−x∣<ϵ.
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,… is said to be Cauchy if ∀ϵ>0,∃N∈Ns.t.∀m,n>N,∣am−an∣<ϵ.
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.