# Value of series, Partialsum?

given is the following series

$$\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$$

And I need to find its value.

How can I start finding it?

Thanks for all

does the Telescop-Summing work here as well?:

$\sum_{n=1}^\infty \frac{1}{4n^2-1}$ now: $\frac{1}{4n^2-1} = \frac{1}{2} * \frac{(2n+1)-(2n-1)}{(2n+1)(2n-1)} = \frac{1}{2} * ( \frac{1}{2n-1} - \frac{1}{2n+1})$

Now I have to "add the sum": $\sum_{n=1}^\infty \frac{1}{4n^2-1} = \frac{1}{2}* [ \sum_{n=1}^\infty \frac{1}{2n-1} - \sum_{n=1}^\infty \frac{1}{2n+1}] = \frac{1}{2} - \frac{1}{4n+2}$ And than for $n \to \infty$ it is $\frac{1}{2}$ ??

HINT: $$\frac{2n+1}{n^2(n+1)^2}=\frac{(n+1)^2-n^2}{n^2(n+1)^2}=\frac1{n^2}-\frac1{(n+1)^2}$$
• Hi, that's a huge hint, thank you very much! :) If I am right, than using the 'Telescoping Sum' in the end I get: $1 - \frac{1}{(n+1)^2}$ and now I look at $n \to \infty$ and I get: The Value is 1. – Vazrael May 21 '13 at 13:47
• Are you suggesting that we go $\frac{2n+1}{n^2(n+1)^2}=\frac{2(n+1)-1}{n^2(n+1)^2}=\frac{2-1}{n^2(n+1)}=\frac{1}{n^2(n+1)}$?...! – user1729 May 21 '13 at 10:52