I am struggling to find the sum of the following series:
$$\sum_{n=1} ^{\infty} \frac{(-1)^n}{(n+1)(n+3)(n+5)}.$$
It seems as though it should be a straightforward telescoping series. I attempted to solve it in the usual way (via partial fractions), but the alternating sign makes the sum so that one cannot cancel out fractions to result in a finite sum of fractions. I know that the series converges by the alternating sign test, and I check on WolframAlpha that the infinite sum converges to $-7/480$. Any thoughts on how to proceed?
Best Answer
We have (through the residue theorem or by linear algebra) : $$ \frac{1}{(n+1)(n+3)(n+5)}=\frac{1}{8}\left(\frac{1}{n+1}-\frac{2}{n+3}+\frac{1}{n+5}\right) $$ as well as (by shifting the summation index): $$ \sum_{n\geq 1}\frac{(-1)^n}{n+1}=-1+\log 2, $$ $$ \sum_{n\geq 1}\frac{(-1)^n}{n+3}=-\frac{5}{6}+\log 2, $$ $$ \sum_{n\geq 1}\frac{(-1)^n}{n+5}=-\frac{47}{60}+\log 2. $$ Just combine them. $\log 2$ cancels out since $1-2+1=0$ (we have a meromorphic function that is $O\left(\frac{1}{|z|^2}\right)$ as $|z|\to +\infty$, hence the sum of its residues is necessary zero).