Answer
$\Sigma_{k=1}^{n} \frac{1}{k(k+1)}=\frac{n}{n+1}$
Work Step by Step
$\Sigma_{k=1}^{n} \frac{1}{k(k+1)}=\Sigma_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)=\Sigma_{k=1}^{n} \frac{1}{k}-\Sigma_{k=1}^{n} \frac{1}{k+1}$
$=1+\frac{1}{2}+\frac{1}{3}+\ldots \ldots+\frac{1}{n}-\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{n+1}\right]$
$=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}-\frac{1}{2}-\frac{1}{3}-\ldots .-\frac{1}{n}-\frac{1}{n+1}$
$=1-\frac{1}{n+1}=\frac{n}{n+1}$
result:
$\Sigma_{k=1}^{n} \frac{1}{k(k+1)}=\frac{n}{n+1}$
