<s>[INST] 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.
 [/INST] $$\begin{eqnarray*}\sum_{n\geq 2}\frac{(-1)^n}{(n+2)(n-1)}&=&\int_{0}^{1}\int_{0}^{1}\sum_{n\geq 2}(-1)^n x^{n+1} y^{n-2}\,dx\,dy\\&=&\int_{0}^{1}\int_{0}^{1}\frac{x^3}{1+xy}\,dy\,dx\\&=&\int_{0}^{1}x^2\log(1+x)\,dx\\&=&\frac{\log 2}{3}-\int_{0}^{1}\frac{x^3}{3(1+x)}\,dx\end{eqnarray*}$$
The last integral is straightforward to compute by writing $x^3$ as $(x^3+1)-1$.
The final outcome is:

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

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