## Calculus (3rd Edition)

By the comparison test, the series $\sum_{n=1}^{\infty}\frac{1}{(1+n)^2}$ coverges.
The p-series $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges and we have $$\frac{1}{1+n}\leq\frac{1}{n}\Longrightarrow \frac{1}{(1+n)^2}\leq\frac{1}{n^2}$$ Then by the comparison test, the series $\sum_{n=1}^{\infty}\frac{1}{(1+n)^2}$ converges.