Problem - 3292
Compute $$\sum_{k=1}^{\infty}\frac{1}{k^2 + k}$$
\begin{align*}
&\sum_{k=1}^{\infty}\frac{1}{k^2 + k} \\
= &\sum_{k=1}^{\infty}\Big(\frac{1}{k}-\frac{1}{k+1}\Big)\\
= &\Big(1 -\frac{1}{2}\Big)+\Big(\frac{1}{2} -\frac{1}{3}\Big)+\Big(\frac{1}{3} -\frac{1}{4}\Big)+\cdots\\
= & 1
\end{align*}