What is the sum of $\sum_{n=1}^\infty\left(\frac{2n+1}{n^4+2n^3+n^2}\right) = $?

Question

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

Answer 1

Since \begin{align*} \frac{2n+1}{n^4+2n^3+n^2}&=\frac{(n+1)^2-n^2}{n^2(n+1)^2}\\ &=\frac 1{n^2}-\frac 1{(n+1)^2} \end{align*}

then we have, for an integer $N>1$ $$\sum_{n=1}^N\left(\frac{2n+1}{n^4+2n^3+n^2}\right)=1-\frac1{(N+1)^2}$$

Answer 2

Hint: $2n+1=(n+1)^2-n^2$, and telescope.

Answer 3

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) )