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
 A: 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. 
A: We have the partial fractions decomposition
$$\frac{1}{k^2+2k}=\frac{1}{2k}-\frac{1}{2k+4}.$$
If you write out the first few terms, you will notice a lot of cancellation:
$$\left(\frac{1}{2}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\left(\frac{1}{6}-\frac{1}{10}\right)+\left(\frac{1}{8}-\frac{1}{12}\right)+\left(\frac{1}{10}-\frac{1}{14}\right)+\cdots+\left(\frac{1}{2n-4}-\frac{1}{2n}\right)+\left(\frac{1}{2n-2}-\frac{1}{2n+2}\right)+\left(\frac{1}{2n}-\frac{1}{2n+4}\right).$$
When the dust clears, all you are left with is
$$\frac{1}{2}+\frac{1}{4}-\frac{1}{2n+2}-\frac{1}{2n+4}$$
which is the same as your partial sum formula.
This also makes it clear why the limit is $3/4$.
