# 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

• spivak 3rd edition page 373 has a nice section on partial fraction decomposition. in general, when you're stuck with an ugly looking rational function, you would use partial fraction decomposition because, when feasible, it guarantees decomposition into simple-looking terms. this is always good when looking for a way to simplify an integral or summation. Commented Jun 16, 2019 at 0:30

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 $$3/4$$.
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.
• Actually, the last $2$ terms in the second term don't cancel, giving $4$ terms in total, like the answer by Thomas Browning shows. Commented Jun 16, 2019 at 0:36