Closed form of an infinite sum $\sum_{k=1}^\infty\frac{1}{k(k+p)}$

Question

I need help on evaluating or finding a closed form for $\sum\limits_{k=1}^{\infty}\frac{1}{k(k+p)}$ for which Wolfram Alpha has given me this solution:

I tried decomposing the fraction but I'm not sure what this yields for me:

$\frac{1}{k(k+p)}$

$= \frac{1}{pk} - \frac{1}{p(k+p)}$

$= \frac{1}{pk} - \frac{1}{pk} + \frac{1}{k(k+p)}$

$= \frac{1}{pk} - \frac{1}{pk} + \frac{1}{pk} - \frac{1}{p(k+p)}$

$= \frac{1}{pk} - \frac{1}{pk} + \frac{1}{pk} - \frac{1}{pk} + \frac{1}{k(k+p)}$

$=\dots$

and so on.