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?
 A: You need to fully expand the partial fraction.
\begin{align}
(-1)^n \over (n+2)(n-1)
&= {((n+2) - (n-1)) \times (-1)^n \over 3\times (n+2)(n-1)} \\
&= {(n+2) \times (-1)^n \over 3\times (n+2)(n-1)} - {(n-1) \times (-1)^n \over 3\times (n+2)(n-1)} \\
&= {(-1)^n \over 3\times (n-1)} - {(-1)^n \over 3\times (n+2)} 
\end{align}
Now this can be handled using Alternating Harmonic Series.
The final result, as Mathematica calculated, is $\frac {-5 + 12 \log 2} {18}$.
A: Writing the $n$th term of your sum in a form that telescopes can be avoid altogether by converting the problem to a double integral as follows.
Noting that
$$\frac{1}{n - 1} = \int_0^1 x^{n - 2} \, dx \qquad \text{and} \qquad \frac{1}{n + 2} = \int_0^1 y^{n + 1} \, dy,$$
the sum can be rewritten as
\begin{align*}
\sum_{n = 2}^\infty \frac{(-1)^n}{n^2 + n - 2} &= \sum_{n = 2}^\infty \frac{(-1)^n}{(n - 1)(n + 2)}\\
&= \int_0^1 \int_0^1 \sum_{n = 2}^\infty (-1)^n x^{n - 2} y^{n + 1} \, dx dy \tag1\\
&= \int_0^1 \int_0^1 \frac{y}{x^2} \sum_{n = 2}^\infty (-xy)^n \, dx dy\\
&= \int_0^1 \int_0^1 \frac{y}{x^2} \cdot \frac{x^2 y^2}{1 + xy} \, dx dy \tag2\\
&= \int_0^1 \int_0^1 \frac{y^3}{1 + xy} \, dx dy\\
&= \int_0^1 y^2 \Big{[} \ln (1 + xy) \Big{]}_0^1 \, dy\\
&= \int_0^1 y^2 \ln (1 + y) \, dy\\
&= \frac{1}{3} \ln (2) - \frac{1}{3} \int_0^1 \frac{y^3}{1 + y} \, dy \tag3\\
&= \frac{1}{3} \ln (2) - \frac{1}{3} \int_0^1 \left (y^2 - y + 1 - \frac{1}{1 + y} \right ) \, dy \tag4\\
&= \frac{1}{3} \ln (2) - \frac{1}{3} \left [\frac{y^3}{3} - \frac{y^2}{2} + y - \ln (1 + y) \right ]_0^1\\
&= \frac{2}{3} \ln (2) - \frac{5}{18}.
\end{align*}
Explanation


*

*Interchanging the summation with the double integration.

*Summing the series which is geometric.

*Integration by parts.

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