compute the sum $$\sum_1^{+\infty}(-1)^n\frac{n}{4n^2-1}$$
I can only see that the summand is odd, hence $$\sum_{-\infty}^{+\infty}(-1)^n\frac{n}{4n^2-1}=0$$
I know there is a standard way of computing summation of series using residue formula, but i can't see how to apply that in this case