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

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 $3/4$.
