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.