Calculate $\sum_{k=0}^{n} \frac{(-1)^kk}{4k^2-1}$

Question

Calculate $\sum_{k=0}^{n} \frac{(-1)^k k}{4k^2-1}$

$\sum_{k=0}^{n} \frac{(-1)^k k}{4k^2-1}=\sum_{k=0}^{n} \frac{(-1)^k k}{(2k-1)(2k+1)}=\sum_{k=0}^{n} (-1)^k k\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})$ after that I get this $\sum_{k=0}^{n} (-1)^k \frac{k}{2k-1}+(-1)^{n+1}\frac{n+1}{2n+1}+\sum_{k=0}^{n}\frac{1}{2} \frac{(-1)^k}{2k+1} $. But that does not help me so much I do not know how to continue, i try everything after I still get the same, can you help me? $\sum_{}^{}$ $\sum_{}^{1}$

Answer 1

$$\sum_{k=0}^n\frac{(-1)^kk}{4k^2-1}=\frac{1}{4}\sum_{k=0}^n(-1)^k\left[\frac{1}{2k-1}+\frac{1}{2k+1}\right]$$ $$=\frac{1}{4}\left[\frac{1}{-1}+\frac{1}{1}-\frac{1}{1}-\frac{1}{3}+\frac{1}{3}+\frac{1}{5}-\cdots+\frac{(-1)^n}{2n-1}+\frac{(-1)^n}{2n+1}\right]$$ $$=\frac{1}{4}\left[\frac{1}{-1}+\frac{(-1)^n}{2n+1}\right]$$

Answer 2

HINT

We have that

$$\sum_{k=0}^n\frac{(-1)^kk}{4k^2-1}=\sum_{k=0}^n\frac{(-1)^k}4\left(\frac{1}{2k+1}+\frac{1}{2k-1}\right)=$$$$=\frac14\left(0\color{red}{-\frac13}-1\color{red}{+\frac15+\frac13-\frac17-\frac15+\frac19+\frac17}-\ldots\right) $$