<s>[INST] How to evaluate $\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2}$? Note: Similar questions have been asked here and here, but this is quite different.
I am trying to evaluate
$$\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2} \quad (1)$$

I re-wrote the fraction as $$ \frac{2n+1}{2n(n+1)^2} = \frac1{2(n+1)} \cdot \frac{2n+1}{n(n+1)}= \frac1{2(n+1)} \left( \frac1n + \frac1{n+1} \right) = \frac1{2} \left( \frac1{n(n+1)} + \frac1{(n+1)^2} \right) = \frac12 \left( \left( \frac1n -\frac1{(n+1)} \right) + \frac1{(n+1)^2} \right)$$
Hence
$$(1) = \frac12 \sum_{n=1}^{\infty}\:  \left( \frac1n -\frac1{(n+1)} \right) + \frac12 \sum_{n=1}^{\infty}\:\frac1{(n+1)^2} = \frac12\lim_{n \to \infty}1-\frac1{n+1}+\frac{\pi^2}{12} = \frac{\pi^2}{12} + \frac12$$
Therefore
$$\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2} = \frac{\pi^2}{12} + \frac12$$

I am unsure about $$ \sum_{n=1}^{\infty}\:\frac1{(n+1)^2} = \frac{\pi^2}{12} $$
We know the basic p-series $$ \sum_{n=1}^{\infty}\:\frac1{n^2} = \frac{\pi^2}{6} $$
Is this solution correct?
 [/INST] 1). $\frac{2n+1}{2n(n+1)^2}=\frac{n+n+1}{2n(n+1)^2}=\frac{1}{2(n+1)^2}+\frac{1}{2n(n+1)}$
So
\begin{align*}
\sum_{n\geq1}{}\frac{2n+1}{2n(n+1)^2}=\frac{1}{2}\sum_{n\geq1}\frac{1}{(n+1)^2}+\frac{1}{2}\sum_{n\geq1}{}\frac{1}{n(n+1)}\\
=\frac{1}{2}(\zeta(2)-1)+\frac{1}{2}\sum_{n\geq1}{}\frac{1}{n}-\frac{1}{n+1}\\
=\frac{1}{2}(\zeta(2)-1)+\frac{1}{2}\\
=\frac{(\pi)^2}{12}\\
\end{align*}
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