How can I find the sum of $\sum_{n=2}^\infty \frac{(-1)^n}{n^2 - n}$ How can I find the sum of the series:
$$\sum_{n=2}^\infty \frac{(-1)^n}{n^2 - n}$$
I did this:
$$\sum_{n=2}^\infty \left(\frac{(-1)^n}{n-1} + \frac{(-1)^{n-1}}{n}\right)$$
But I am not sure how to proceed.
You are on the right track:
$$\sum_{n=2}^\infty \frac{(-1)^n}{n^2 - n}=\sum_{n=2}^\infty \left(\frac{(-1)^n}{n-1} + \frac{(-1)^{n-1}}{n}\right).$$
Now note that
$$\ln(2)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\sum_{n=2}^{\infty}\frac{(-1)^{n}}{n-1}.$$
$$
\sum_{n=2}^\infty \frac{(-1)^n}{n^2-n}=
\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)}=\lim_{x\to 1^{-}}
\sum_{n=1}^\infty \frac{(-1)^{n+1}x^{n+1}}{n(n+1)}
=\lim_{x\to 1^{-}}\int_0^x\sum_{n=1}^\infty \frac{(-1)^{n+1}t^n}{n}\,dt \\=\lim_{x\to 1^{-}}\int_0^x \log (t+1)\,dt=\lim_{x\to 1^{-}}\big((x+1)\log (x+1)-x\big)=2\log 2-1
$$
Note. The second "=" is a consequence of Abel's Theorem.