Franklin Pezzuti Dyer’s Blog

Telescoping Sums

2017 June 9

Find the values of each of the following infinite sums:
\[\sum_{x=1}^\infty \frac{1}{x(x+1)}\]
\[\sum_{x=2}^\infty \frac{1}{x^2-1}\]
\[\sum_{x=1}^\infty \frac{1}{x(x+1)(x+2)}\]
\[\sum_{x=1}^\infty \frac{x}{2^x}\]

Okay, let’s start with the first problem:
\[\sum_{x=1}^\infty \frac{1}{x(x+1)}\]
This is a simple telescoping sum problem. Note that
\[\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}\]
So, when expanded, the infinite sum looks like
\[\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...\]
Look! Each term other than the very first one forms a zero pair with some other term. They all cancel out, and we are left with
\[\sum_{x=1}^\infty \frac{1}{x(x+1)}=1\]
as our answer. But wait a minute… couldn’t we do an infinite sum like
\[\sum_{x=1}^\infty \frac{1}{x(x+2)}\]
the same way? Since
\[\frac{1}{x(x+2)}=\frac{1}{2}\bigg(\frac{1}{x}-\frac{1}{x+2}\bigg)\]
then the entire sum is
\[\frac{1}{2}\bigg(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...\bigg)\]
Now the only terms that don’t get cancelled are \(1\) and \(\frac{1}{2}\), so the sum is
\[\sum_{x=1}^\infty \frac{1}{x(x+2)}=\frac{1}{2}\bigg(1+\frac{1}{2}\bigg)=\frac{3}{4}\]
In fact, we can generalize this method of solving sums of this sort. If we have a sum of the form
\[\sum_{x=1}^\infty \frac{1}{x(x+k)}\]
We can use partial fractions to set
\[\frac{1}{x(x+k)}=\frac{1}{k}\bigg(\frac{1}{x}-\frac{1}{x+k}\bigg)\]
And so, by expanding, we can see that the sum becomes
\[\frac{1}{k}\bigg(1-\frac{1}{k+1}+\frac{1}{2}-\frac{1}{k+2}+...+\frac{1}{k}+\frac{1}{2k}+\frac{1}{k+1}+\frac{1}{2k+1}+...\bigg)\]
Which can be rearranged to form
\[\frac{1}{k}\bigg(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k}+\frac{1}{k+1}-\frac{1}{k+1}+\frac{1}{k+2}-\frac{1}{k+2}+...\bigg)\]
All of the terms cancel except for those from \(1\) to \(k\), so we get
\[\frac{1}{k}\bigg(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k}\bigg)\]
and we have the formula
\[\sum_{x=1}^\infty \frac{1}{x(x+k)}=\frac{1}{k}\sum_{x=1}^k \frac{1}{x}\]
or
\[\sum_{x=1}^\infty \frac{1}{x(x+k)}=\frac{1}{k}H_k\]
where \(H_n\) denotes the nth harmonic number.
Okay! Onto the next sum:
\[\sum_{x=2}^\infty \frac{1}{x^2-1}\]
We can split up the fraction into
\[\frac{1}{(x+1)(x-1)}\]
\[\frac{1}{2}\bigg(\frac{1}{x-1}-\frac{1}{x+1}\bigg)\]
And so, when we expand the sum, we get
\[\frac{1}{2}\bigg(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...\bigg)\]
\[\frac{1}{2}\bigg(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...\bigg)\]
\[\frac{1}{2}\bigg(1+\frac{1}{2}\bigg)\]
\[\frac{3}{4}\]
and so we have our answer:
\[\sum_{x=2}^\infty \frac{1}{x^2-1}=\frac{3}{4}\]

Are you noticing a pattern in these summation problems?

Suppose we have any summation in the form
\[\sum_{x=1}^\infty f(x)-f(x+k)\]
or
\[\lim_{n\to \infty}\sum_{x=1}^n f(x)-f(x+k)\]
we can start by finding a formula for
\[\sum_{x=1}^n f(x)-f(x+k)\]
which can be found by expanding the sum and telescoping it:
\[f(1)-f(k+1)+f(2)-f(k+2)+...+f(n)-f(k+n)\]
If we assume that \(n \gt k\), then we have
\[f(1)-f(k+1)+f(2)-f(k+2)+...+f(k)-f(2k)+f(k+1)-f(2k+1)+...+f(n)-f(k+n)\]
Which collapses to
\[f(1)+f(2)+...+f(k)-f(n+1)-f(n+2)-...-f(n+k)\]
and this is equal to
\[\sum_{x=1}^k f(x)-\sum_{x=n+1}^{n+k} f(x)\]
And our original infinite sum is
\[\lim_{n\to \infty}\Bigg(\sum_{x=1}^k f(x)-\sum_{x=n+1}^{n+k} f(x)\Bigg)\]
If \(\lim_{x\to\infty}f(x)\) exists, then we can say that this is equal to
\[\sum_{x=1}^k f(x)-\lim_{x\to\infty}kf(x)\]
This is the formula for a telescoping sum:
\[\sum_{x=1}^\infty f(x)-f(x+k)=\sum_{x=1}^k f(x)-\lim_{x\to\infty}kf(x)\]
Given that \(k\) is a natural number.

Moving on to the next problem:
\[\sum_{x=1}^\infty \frac{1}{x(x+1)(x+2)}\]
Using partial fractions, we can split up the fraction to get
\[\frac{1}{x(x+1)(x+2)}=\frac{1}{2}\frac{1}{x}-\frac{1}{x+1}+\frac{1}{2}\frac{1}{x+2}\]
\[\frac{1}{x(x+1)(x+2)}=\frac{1}{2}\bigg(\frac{1}{x}-\frac{1}{x+1}\bigg)-\frac{1}{2}\bigg(\frac{1}{x+1}-\frac{1}{x+2}\bigg)\]
and so our sum is equal to
\[\frac{1}{2}\sum_{x=1}^\infty \bigg(\frac{1}{x}-\frac{1}{x+1}\bigg)-\frac{1}{2}\sum_{x=1}^\infty \bigg(\frac{1}{x+1}-\frac{1}{x+2}\bigg)\]
\[\frac{1}{2}\bigg(1-\frac{1}{2}\bigg)\]
\[\frac{1}{4}\]
And so the answer to this problem is
\[\sum_{x=1}^\infty \frac{1}{x(x+1)(x+2)}=\frac{1}{4}\]

Next up is a different type of sum:
\[\sum_{x=1}^\infty \frac{x}{2^x}\]
Yay! Something that can’t be done using partial fractions! This one telescopes in a different way. Look what happens if we multiply and divide the sum by \(\frac{1}{2}\):
\[\frac{\frac{1}{2}}{\frac{1}{2}}\sum_{x=1}^\infty \frac{x}{2^x}\]
\[\frac{(1-\frac{1}{2})}{\frac{1}{2}}\sum_{x=1}^\infty \frac{x}{2^x}\]
\[\frac{1}{\frac{1}{2}}\sum_{x=1}^\infty (1-\frac{1}{2})\frac{x}{2^x}\]
\[2\sum_{x=1}^\infty \bigg(\frac{x}{2^x}-\frac{x}{2^{x+1}}\bigg)\]
Now look what happens when we write this out:
\[2\bigg(\frac{1}{2}-\frac{1}{4}+\frac{2}{4}-\frac{2}{8}+\frac{3}{8}-\frac{3}{16}+...\bigg)\]
\[2\bigg(\frac{1}{2}+\frac{-1+2}{4}+\frac{-2+3}{8}+...\bigg)\]
\[2\bigg(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...\bigg)\]
Now we have a geometric series inside of the parentheses. That geometric series can be easily calculated to converge to \(1\) using the formula, and we have
\[2(1)\]
\[2\]
and so
\[\sum_{x=1}^\infty \frac{x}{2^x}=2\]

Great. All of these sums were telescoping sums, but in a future post I hope to tackle a few infinite sums that cannot be solved in this way.