<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] As has been already pointed out in the comments,
$$\sum_{n = 1}^\infty \frac{1}{(n+1)^2} = \frac{1}{4} + \frac 19 + ... = \frac{\pi^2}{6}-1$$
Therefore, your solution would be
$$\frac{1}{2}\left( \sum_{n=1}^\infty \frac 1n - \frac {1}{n+1} + \sum_{n=1}^\infty\frac{1}{(n+1)^2}\right) = \frac 12 \left( 1 + \frac{\pi^2}{6} - 1\right) = \boxed{\frac{\pi^2}{12}}$$
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