# Calculate $\sum_{n=1}^\infty \frac{n+1}{n(n+2)^2}$ using Basel Problem sum

I have this series I need to calculate: $$\sum_{n=1}^\infty \frac{n+1}{n(n+2)^2}$$ I've been trying to simplify the fraction via partial fraction decomposition, and I know I somehow need to make use of the Basel Problem, meaning that $$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$ , but I don't know how to adress the movement of the index $$n$$ by $$2$$ in my original problem.

Any tips of how to proceed from here?

• Don't know if it helps, but what about $$\sum_{i=2}^\infty {\frac{i}{(i-1)(i+1)^2}}=\sum_{i=2}^\infty {\frac{i}{i^3+i^2-i-1}}$$ – Dr. Mathva Mar 31 '19 at 9:24

Write (for example using Partial-fraction decomposition) $$\frac{n+1}{n(n+2)^2} = \frac{1}{2(n+2)^2}+\frac{1}{4}\left(\frac{1}{n}-\frac{1}{n+2}\right).$$ The second bracket summed is a telescoping sum, specifically $$\sum_{n=1}^k\frac{1}{n}-\frac{1}{n+2}=\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\dots+\frac{1}{k}-\frac{1}{k+2},$$ which after canceling terms gives $$\sum_{n=1}^k\frac{1}{n}-\frac{1}{n+2}=1+\frac{1}{2}-\frac{1}{k+1}-\frac{1}{k+2} \to \frac{3}{2}\,\, \text{as }k\to \infty.$$
For the first expression, notice $$\sum_{n=1}^\infty\frac{1}{(n+2)^2}=\sum_{n=3}^\infty\frac{1}{n^2}=\sum_{n=1}^\infty\frac{1}{n^2}-1-\frac{1}{4}.$$ Can you put these together and finish?