## Telescoping Sums

2017 June 9

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

$\sum_{x=1}^\infty \frac{1}{x(x+1)}$
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.