Not Riemann Zeta Function

$\sum\limits_{i=1}^{\infty} \frac{2i+1}{i^2(i+1)^2}$ =

$\sum\limits_{i=1}^{\infty} \frac{i^2+2i+1}{i^2(i+1)^2}-\frac{i^2}{i^2(i+1)^2}$ =

$\sum\limits_{i=1}^{\infty} \frac{(i+1)^2}{i^2(i+1)^2}-\frac{i^2}{i^2(i+1)^2}$ =

$\sum\limits_{i=1}^{\infty} \frac{1}{i^2} - \frac{1}{(i+1)^2}$ . Notice that in our next summation the next term is going to be equal to our last term here and this pattern keeps going till infinity.Thus the value is :

$\sum\limits_{i=1}^{\infty} \frac{1}{i^2} =$

$\sum\limits_{i=1}^{\infty} \frac{1}{1^2}=\boxed{1}$ .

it is simple $\sum_{n=1}^\infty (\frac{2n+1}{(n(n+1))^2})$ $\sum_{n-1}^\infty (\frac{1}{n(n+1)}\times \frac{2n+1}{n(n+1)})$ $\sum_{n-1}^\infty (\frac{1}{n}-\frac{1}{n+1})(\frac{1}{n}+\frac{1}{n+1})$ we know that $(a-b)(a+b) =a^2 -b^2$ so, $\sum_{n-1}^\infty (\frac{1}{n^2}-\frac{1}{(n+1)^2})$ the values are $1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}......\infty$ arrange them to $1+( -\frac{1}{4}+\frac{1}{4})+( -\frac{1}{9}+\frac{1}{9}).......\infty$ we get $\boxed{1}$