A calculus problem by Ben Habeahan

The sum $S$ is:

$\begin{aligned} S & = \sum_{k=1}^\infty \frac{k+1}{k^2(k+2)^2} \quad \quad \quad \quad \small \color{#3D99F6} {\left[\text{We note that }\frac{1}{k^2} - \frac{1}{(k+2)^2} = \frac{4(k+1)}{k^2(k+2)^2}\right]} \\ & = \frac{1}{4} \sum_{k=1}^\infty \left( \frac{1}{k^2} - \frac{1}{(k+2)^2} \right) \\ & = \frac{1}{4} \left( \sum_{k=1}^\infty \frac{1}{k^2} - \sum_{k=1}^\infty \frac{1}{(k+2)^2} \right) \\ & = \frac{1}{4} \left( \sum_{k=1}^\infty \frac{1}{k^2} - \sum_{k=3}^\infty \frac{1}{k^2} \right) \\ & = \frac{1}{4} \left( \frac{1}{1^2} + \frac{1}{2^2} \right) = \frac{1}{4} \left( 1 + \frac{1}{4} \right) = \frac{5}{16} \end{aligned}$

$\Rightarrow A + B = 5 + 16 = \boxed{21}$ .

General term is => (n+1)/{ n^2 * (n+2)^2 } which is same as => { 1/n^2 - 1/(n+2)^2 } * (1/4)

For k = 1 to ... only two terms remains which do not get cancelled by any other.. ={ 1/1 + 1/4 } * 1/4 =5/16

so A+B = 21