Thank you to Surb for that hint.
$$\sum_{n=1}^{k}\frac{1}{n^2+n}=\sum_{n=1}^{k}\frac{1}{n}-\frac{1}{n+1}$$ $$=\sum_{n=1}^{k}\frac{1}{n}-\sum_{n=1}^{k}\frac{1}{n+1}$$ $$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}...+\frac{1}{k-1}+\frac{1}{k}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; -(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k-1}+\frac{1}{k}+\frac{1}{k+1})$$$$\begin{align*} \sum_{n=1}^{k}\frac{1}{n^2+n}&=\sum_{n=1}^{k}\frac{1}{n}-\frac{1}{n+1}\\ &=\sum_{n=1}^{k}\frac{1}{n}-\sum_{n=1}^{k}\frac{1}{n+1} \\ &=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k-1}+\frac{1}{k}\\\\ &\phantom{\frac{1}{1}+}-\Big(\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k-1}+\frac{1}{k}+\frac{1}{k+1}\Big)\end{align*}$$
Everything cancels out except for $\frac{1}{1}-\frac{1}{k+1}=\frac{k}{k+1}$.
