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 $\frac{1}{2}$ ??

split it up the numerator to 2n + 2 - 1 then factor to 2(n+1) -1.
By spliting the fraction you can cancel terms.
This should allow you to see this is a finite result.
