sum $\displaystyle \sum _{n=2}^{\infty}\frac{(-1)^n}{n^2+n-2}$ I have the following series:
$$\sum _{n=2}^{\infty}\frac{(-1)^n}{n^2+n-2}$$
I am not able to do the telescoping process in the above series. I converted it into the following partial fraction:
$$\sum _{n=2}^{\infty}\frac{(-1)^n}{(n+2)(n-1)}$$
But nothing seems to cancel (as usually happens) using the telescoping method. How can I solve the above series? Is there any other method to do the above problem?

$$\sum_{k=2}^{+\infty}\frac{(-1)^n}{n^2+n-2}=\frac{1}{3}\sum_{k=2}^{+\infty}(-1)^n\left(\frac{1}{n-1}-\frac{1}{n+2}\right)=$$
$$=\frac{1}{3}\left(1-\frac{1}{4}-\frac{1}{2}+\frac{1}{5}+\frac{1}{3}-\frac{1}{6}-\frac{1}{4}+\frac{1}{7}+...\right)=$$
$$=-\frac{1}{3}\left(1-\frac{1}{2}+\frac{1}{3}\right)+\frac{2}{3}\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k}=\frac{2}{3}\ln2-\frac{5}{18}.$$
