Infinite sum of alternating telescoping series 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?
 A: 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).
A: Rewrite the alternating sum as a difference of two infinite sums. $$\sum_{n=1} ^{\infty} \frac{(-1)^n}{(n+1)(n+3)(n+5)} = \sum_{n=1} ^{\infty} \frac{1}{(2n+1)(2n+3)(2n+5)}-\sum_{n=1} ^{\infty} \frac{1}{2n(2n+2)(2n+4)}$$ You'll probably wnat to convince yourself of the equality. Now you can use partial fraction decomposition to decompose $\frac{1}{(2n+1)(2n+3)(2n+5)}$ and $\frac{1}{2n(2n+2)(2n+4)}$ into three separate fractions each, giving you six infinite sums in total. From there you should be able to make some judicious cancellations and get your result. You may find one of my previous questions helpful at this point. Exact value of $\sum_{n=1}^\infty \frac{1}{n(n+k)(n+l)}$ for $k \in \Bbb{N}-\{0\}$ and $l \in \Bbb{N}-\{0,k\}$
A: Partial Decomposition provides:
$S = \sum_{1}^{\infty} (-1)^n[\frac{1}{8(n+1)} - \frac{1}{4(n+3)}+\frac{1}{8(n+5)}]$
Splitting S_Even and S_Odd
$S_{Even}  = \frac{1}{8.3} - \frac{1}{4.5}+\frac{1}{8.5}$
$S_{Odd} = -\frac{1}{8.2} +\frac{1}{4.4} - \frac{1}{8.4}$
Everything else cancels out
When you sum these you get $S = \dfrac{-7}{480}$
Provided I have not made any calculation error
