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
So, when expanded, the infinite sum looks like
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
then the entire sum is
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
And so, by expanding, we can see that the sum becomes
Which can be rearranged to form
All of the terms cancel except for those from \(1\) to \(k\), so we get
and we have the formula
\[\sum_{x=1}^\infty \frac{1}{x(x+k)}=\frac{1}{k}\sum_{x=1}^k \frac{1}{x}\]
\[\sum_{x=1}^\infty \frac{1}{x(x+k)}=\frac{1}{k}H_k\]
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
And so, when we expand the sum, we get
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
\[\sum_{x=1}^\infty f(x)-f(x+k)\]
\[\lim_{n\to \infty}\sum_{x=1}^n f(x)-f(x+k)\]
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:
If we assume that \(n \gt k\), then we have
Which collapses to
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
and so our sum is equal to
\[\frac{1}{2}\sum_{x=1}^\infty \bigg(\frac{1}{x}-\frac{1}{x+1}\bigg)-\frac{1}{2}\sum_{x=1}^\infty \bigg(\frac{1}{x+1}-\frac{1}{x+2}\bigg)\]
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:
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
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.