Looking to the following series:
$$\sum_{n=1}^\infty \frac{(-1)^n(4n)}{4n^2-1}$$
It converges according to Leibniz criteria. However it does not seem to be a telescopic series (if you take partial fractions, you end up with two positive terms), neither a geometric one, so I can not figure out a way to find its sum. Maybe I am missing something here.
Thanks for your time and I appreciate any help.
Best Answer
It is in fact telescopic. For the partial sum: $$ S_N = \sum_{n=1}^N (-1)^n \left( \frac{1}{2n-1}+\frac{1}{2n+1}\right)= \sum_{n=1}^N \frac{(-1)^n}{2n-1} - \sum_{k=2}^{N+1} \frac{(-1)^k}{2k-1} =-1+\frac{(-1)^N}{2N+1}\rightarrow -1$$