How to compute $\sum_{k=1}^{\infty}{\frac{1}{k^2+2k}}$? To whom this may concern,
i am struggling with partial sum formulas. I don't really get why you would need to perform a partial fraction decomposition or how you know that you have to.
I started by getting trying to get a grasp of the series:$$\sum_{k=1}^{n}{\frac{1}{k^2+2k}}=\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\cdots + \frac{1}{n(n+2)}$$
But the partial sum formula is, according to WolframAlpha, $\sum_{k=1}^{n}{\frac{1}{k^2+2k}}=\frac{3n^2+5n}{4(n+1)(n+2)}$ HOW?????? I beg you to be as detailed as possible, i really want to understand WHY

Note that $\frac 1 {k^{2}+2k} =\frac 1 2(\frac 1 k -\frac 1 {k+2})$. In $(1+\frac 1 2+...+\frac 1 n) -(\frac 1 3+...+\frac 1 {n+2})$  all terms except the first two in the first term and the last two  terms in the second term cancel. Can you now compute the partial sum. The answer given in WolframAlfa is correct. 
