A telescoping sum is a sum
of differences. We look at a general example. If $a_n
= f(n)-f(n+1)$, then $$\sum_{n=1}^\infty a_n =\sum_{n=1}^\infty
\left(f(n)-f(n+1)\right)=
\left(f(1)-f(2)\right)+\left(f(2)-f(3)\right)+\left(f(3)-f(4)\right)
+ \cdots.$$DO: Carefully find
the $n^{th}$ partial sum of this series. It is a
telescoping sum. $\displaystyle s_n=\sum_{i=1}^n
\left(f(i)-f(i+1)\right) =
\left(f(1)-f(2)\right)+\left(f(2)-f(3)\right) +
\cdots+(f(n-1)-f(n))+(f(n)-f(n+1))$. You can cancel many
terms (Do this.) to be left
with only the two terms $s_n=f(1)-f(n+1)$. Thus we can
compute the convergence/divergence of the series, and if it
converges, we can find its value: $$\sum_{n=1}^\infty
\left(f(n)-f(n+1)\right)=\lim_{n\to\infty}s_n=\lim_{n\to\infty}(f(1)-f(n+1))=f(1)-\lim_{n\to\infty}f(n+1).$$A series with telescoping partial sums is one
of the rare series with which we can compute the value of the
series by using the definition of a series as the limit of its
partial sums.

Example: $\displaystyle\sum_{n=2}^\infty \frac{1}{n(n-1)} =
\sum_{n=2}^\infty\left( \frac{1}{n-1}-\frac{1}{n}\right)$. DO: Check this equality by using partial fraction decomposition on $\displaystyle\frac{1}{n(n-1)}$, then write out
the $n^{th}$ partial sum of this series and cancel terms.

$\displaystyle s_n=\sum_{i=2}^n\left(
\frac{1}{i-1}-\frac{1}{i}\right)=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots+\left(\frac{1}{n-2}-\frac{1}{n-1}\right)+\left(\frac{1}{n-1}-\frac{1}{n}\right)$
$\displaystyle=1-\frac{1}{n}$ DO: find the value of the series
by computing the limit of the partial sums.

$\displaystyle\lim_{n\to\infty}s_n=\lim_{n\to\infty}\left(1-\frac{1}{n}\right)=1-0=1$.
So $\displaystyle\sum_{n=2}^\infty \frac{1}{n(n-1)}=1$.
.

This video will play with these ideas in a more abstract way, and
will tie them to integrals, derivatives, and the FTC.
Depending upon your instructor, the information in the video, while
interesting, is optional.