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To calculate an infinite sum exactly, try antidifferencing

Quick description

If you are trying to calculate the sum \sum_{n=1}^\infty a_n, then sometimes it is possible to spot another sequence (b_n) such that a_n=b_n-b_{n+1} and b_n\rightarrow 0. If that is the case, then your sum is equal to b_1.


The definition of an infinite sum.

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Example 1

An infinite sum that is well known to be straightforward to calculate exactly is the sum

 \frac 1{1\cdot 2}+\frac 1{2\cdot 3}+\frac 1{3\cdot 4}+\dots

The usual technique for summing this is to observe that \frac 1{n(n+1)}=\frac 1n-\frac 1{n+1}, from which it follows that

 \frac 1{1\cdot 2}+\frac 1{2\cdot 3}+\dots+\frac 1{N(N+1)}=1-\frac 1{N+1},

which tends to 1 as N tends to infinity.

This points to a circumstance in which an infinite sum can be evaluated exactly: we can work out the discrete analogue of an "antiderivative": that is, we have a sequence (a_n) and can spot a nice sequence (b_n) such that b_n-b_{n+1}=a_n. Of course, b_n must in that case be the partial sum a_1+\dots+a_n (up to an additive constant). So is this really making the more or less tautologous observation that if you have a nice formula for the partial sums and can see easily what they converge to then you are done?

It isn't quite, because we could have spotted that \frac 1{n(n+1)}=\frac 1n-\frac 1{n+1} without working out any partial sums.

General discussion

Let us try to understand better why the above trick is not a universal method for calculating all infinite sums, by contrasting the above example with the example of the sum

 \frac 1{1^2}+\frac 1{2^2}+\frac 1{3^2}+\dots

Can we find some sequence (b_n) such that b_n-b_{n+1}=1/n^2? It seems difficult just to spot such a sequence, so instead let us try to be systematic about it. It's not quite clear how to pick b_1, so let us set it to be \sigma. Then \sigma-b_2=1, so b_2=\sigma-1. Next, b_2-b_3=1/4, so b_3=\sigma-1-1/4. In general, we find that b_n=\sigma-1-1/4-\dots-1/(n-1)^2. Since we also want b_n to tend to zero, this tells us what \sigma must be: \sum_{n=1}^\infty 1/n^2, precisely the sum we were trying to calculate!

The reason the technique worked in the example above was that the partial sums turned out to have a nice formula.

Remark A simple method of generating undergraduate mathematics exercises is to reverse the telescoping-sums idea in order to create infinite sums that can be evaluated exactly. For example, take the sequence b_n=1/(1+n^2). This tends to 0 as n tends to infinity. Now let
a_n=b_n-b_{n+1}=\frac 1{1+n^2}-\frac 1{1+(n+1)^2}=\frac{(n+1)^2-n^2}{(1+n^2)(1+(n+1)^2}=\frac{2n+1}{(n^2+1)(n^2+2n+2)}.
The question you then ask is to calculate the sum \sum_{n=1}^\infty\frac{2n+1}{(n^2+1)(n^2+2n+2)}. A smart student will use partial fractions to discover that this is a telescoping sum and will end up with the answer b_1=1/2.

If this method works, it is a bit like managing to calculate an integral by antidifferentiating the integrand.


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