Find the sum of series $\sum\limits_{n=2}^{+\infty}\frac{(-1)^n}{n^2+n-2}$. Using the power series $\sum\limits_{n=2}^{+\infty}\frac{1}{n^2+n-2}x^{n}$, the first derivative gives $\sum\limits_{n=2}^{+\infty}\frac{n}{n^2+n-2}x^{n-1}$ where the sum doesn't exist.
Integration of function $\frac{1}{n^2+n-2}x^n$ given $\frac{1}{(n+1)(n^2+n-2)}x^{n+1}$, and after forming the series, $\sum\limits_{n=2}^{+\infty}\frac{1}{(n+1)(n^2+n-2)}x^{n+1}$. After multiple integrations, I didn't found any simplifications.
 A: Decompose $n^2+n-2$ as $(n+2)(n-1)$ and then do: $$\frac{1}{n^2+n-2}=\frac{A}{n+2}+\frac{B}{n-1} \Rightarrow B=-A=\frac{1}{3}$$ And from this you can make the sum (just putting 1/3 out of the same as it is a common factor): $$\frac{1}{3}\left(\sum_{n=2}^{\infty}(-1)^n \left(\frac{1}{n-1} - \frac{1}{n+2}\right)\right) $$ Now, as you can guess, the sum of the sum of two fractions is the just the sum of the first plus the sum of the second. Consider the first one: $$\sum_{n=2}^{\infty}\frac{(-1)^n }{n-1}$$ now put n=2, the first sum, we begin with a one in the denominator and a plus one in the denominator, so let's do this: $$\sum_{n=2}^{\infty}\frac{(-1)^n }{n-1} = \sum_{n=2}^{\infty} (-1) \cdot\frac{(-1)^{n-1}}{n-1} = (-1) \sum_{n=2}^{\infty} \cdot\frac{(-1)^{n-1}}{n-1} = -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}$$ Notice that the last equality is true (plug the first values and see it for yourself) Now, I did the same to the second one and obtained the sum: $$\sum_{n=2}^{\infty}\frac{(-1)^n }{n+2} = \frac{1}{4} - \frac{1}{5} +... = \left(-1+\frac{1}{2} -\frac{1}{3} +\frac{1}{4} -\frac{1}{5} \right) + \left(1-\frac{1}{2}+\frac{1}{3} \right) = $$ $$= \sum_{n=1}^{\infty} \frac{(-1)^n}{n} +1 -\frac{1}{2} + \frac{1}{3} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n} +\frac{5}{6}$$
Now putting it all together we get: 
$$\frac{1}{3}\left(-\frac{5}{6} -2\sum_{n=1}^{\infty}\frac{(-1)^n}{n} \right)$$
Now that last sum is simply $-\log(2)$ therefore the total sum is: $$\sum_{n=2}^{\infty} \frac{(-1)^n}{n^2+n-2} = \frac{1}{3}\left(2\log(2)-\frac{5}{6}\right)$$
A: $$\begin{eqnarray*}\sum_{n\geq 2}\frac{(-1)^n}{(n+2)(n-1)}&=&\int_{0}^{1}\int_{0}^{1}\sum_{n\geq 2}(-1)^n x^{n+1} y^{n-2}\,dx\,dy\\&=&\int_{0}^{1}\int_{0}^{1}\frac{x^3}{1+xy}\,dy\,dx\\&=&\int_{0}^{1}x^2\log(1+x)\,dx\\&=&\frac{\log 2}{3}-\int_{0}^{1}\frac{x^3}{3(1+x)}\,dx\end{eqnarray*}$$
The last integral is straightforward to compute by writing $x^3$ as $(x^3+1)-1$.
The final outcome is:

$$\sum_{n\geq 2}\frac{(-1)^n}{(n+2)(n-1)} = \color{red}{-\frac{5}{18}+\frac{\log(4)}{3}}.$$

