Sum of the series: $\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}$ is? This was a question I confronted in JAM 2016. I tried the following steps: 
$$\displaystyle\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2} 
\implies\sum_{n=2}^\infty\frac{(-1)^n}{(n-1)(n+2)}\implies\sum_{n=2}^\infty(-1)^n\frac{1}{3}\cdot[\frac{1}{(n-1)}-\frac{1}{(n-2)}]$$$$\implies\lim_
{n\to\infty}(-1)^n\cdot\frac{1}{3}[1-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}+...] $$
From the options it was evident that the sum had a relation with $\log_{e}2$, however I couldn't manage to solve the problem. What was I missing?
 A: $$\frac{1}{3}\sum_{n=2}^{+\infty}(-1)^n\left(\frac{1}{n-1}-\frac{1}{n+2}\right)=$$
$$=\frac{1}{3}\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}+\frac{1}{3}\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}-\frac{1}{3}\left(1-\frac{1}{2}+\frac{1}{3}\right)=$$
$$=\frac{1}{3}\ln2+\frac{1}{3}\ln2-\frac{1}{3}\left(1-\frac{1}{2}+\frac{1}{3}\right)=\frac{2}{3}\ln2-\frac{5}{18}.$$
A: $$S=\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}$$
Change the indice of the sum
$$m=n-1$$
$$
\begin {align}
S=&-\sum_{m=1}^\infty\frac{(-1)^m}{m(m+3)} \\
S=&-\frac 1 3\sum_{m=1}^\infty{(-1)^m}\left ({\frac 1 m -\frac 1 {m+3}} \right ) \\
S=&\frac 1 3 \ln 2 +\frac 1 3 \left ( \sum_{m=1}^\infty{\frac {(-1)^m} {m+3}} \right )  \\
S=&\frac 1 3 \ln 2 -\frac 1 3 \left ( \sum_{m=4}^\infty{\frac {(-1)^m} {m}} \right ) \\
S=&\frac 2 3 \ln 2 +\frac 1 3 \left ( \sum_{m=1}^3{\frac {(-1)^m} {m}} \right ) \\
S=&\frac 2 3 \ln 2 +\frac 1 3 \left ( -1+\frac 12-\frac 13\right )
\end{align}
$$
Finally
$$\boxed {S=\frac 2 3 \ln 2 -\frac 5{18}}$$ 
A: Like Steven Sadnicki said, you factored out the $(-1)^n$ term even though it's inside the sum, so the summands are of alternating sign. Here's a solution:
$$\sum_{n=2}^{\infty} \frac{(-1)^n}{n^2 + n - 2 } = \sum_{n=2}^{\infty} \frac{(-1)^n}{(n-1)(n+2)} = \sum_{n=2}^{\infty} (-1)^n [ \frac{1}{3} \cdot ( \frac{1}{n-1} - \frac{1}{n+2})] = \frac{1}{3} \cdot \lim_{n \to \infty} (1 - \frac{1}{4} - \frac{1}{2} + \frac{1}{5} + \frac{1}{3} - \frac{1}{6} - \frac{1}{4} + \frac{1}{7} + \frac{1}{5} - \dots) = \frac{1}{3} \lim_{n \to \infty}(1 - \frac{1}{2} + \frac{1}{3} + 2\sum_{n=4}^{\infty}\frac{(-1)^{n-1}}{n}) = \dots = -\frac{5}{18} - \frac{2}{3}\ln{2}.$$
