Sum of a Telescoping Series
Consider the series defined by
\[a_k = \frac{3}{k(k+2)}\]
.We can write
\[a_k = \frac{1.5}{k}-\frac{1.5}{k+2}\]
.The sum of the series is
\[\begin{equation} \begin{aligned} \sum^n_{k=1} a_k &= a_1+a_2+a_3+a_4+...+a_{n-3}+a_{n-2}+a_{n-1}+a_n \\ &= (\frac{1.5}{1}-\frac{1.5}{3})+(\frac{1.5}{2}-\frac{1.5}{4}) \\ &+ (\frac{1.5}{3}-\frac{1.5}{5})+(\frac{1.5}{4}-\frac{1.5}{6}) \\ &+ ...+(\frac{1.5}{n-3}-\frac{1.5}{n-1})+(\frac{1.5}{n-2}-\frac{1.5}{n}) \\ &+ (\frac{1.5}{n-1}-\frac{1.5}{n+1})+(\frac{1.5}{n}-\frac{1.5}{n+2}) \\ &= \frac{1.5}{1}+ \frac{1.5}{2}-\frac{1.5}{n+1} - \frac{1.5}{n+2} \\ &= \frac{9n^2+3n-9}{4n^2+12n+8}\end{aligned} \end{equation}\]
If
\[n= \infty\]
then the highest powers of \[n\]
dominate, so \[S_{\infty}=\frac{9}{4}\]
. 
