What is the sum of $\sum_{n=1}^\infty\left(\frac{2n+1}{n^4+2n^3+n^2}\right) = $? $$\sum_{n=1}^\infty\left(\frac{2n+1}{n^4+2n^3+n^2}\right)=\sum_{n=1}^\infty\left(\frac{2n+1}{n^2}\frac1{{(n+1)}^2}\right)$$
I assume that I should get a telescoping sum in some way, but I'm couldn't find it yet.

You'll get a telescoping series, but that is likely just only one of the ingredients for a solution. First, I'll separate the numerator in two parts:
$$
\frac{2n+1}{n^2}\frac{1}{(n+1)^2}=\frac{n+n+1}{n^2(n+1)^2}
$$
$$
\frac{2n+1}{n^2}\frac{1}{(n+1)^2}=\frac{1}{n(n+1)^2}+\frac{1}{n^2(n+1)}
$$
And of course your telescoping element:
$$
\frac{1}{n(n+1)}=\left( \frac{1}{n}-\frac{1}{n+1}\right)
$$
Therefore:
$$
\frac{2n+1}{n^2}\frac{1}{(n+1)^2}=\frac{1}{n(n+1)}-\frac{1}{(n+1)^2}+\frac{1}{n^2}-\frac{1}{n(n+1)}
$$
$$
\frac{2n+1}{n^2}\frac{1}{(n+1)^2}=-\frac{1}{(n+1)^2}+\frac{1}{n^2}
$$
And the result is simply 1. As would [WolframAlpha confirm ]( https://www.wolframalpha.com/input/?i=sum((2*n%2B1)%2F((n%5E2(n%2B1)%5E2) )
