sum $\displaystyle \sum _{n=2}^{\infty}\frac{(-1)^n}{n^2+n-2}$

Question

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?

Answer 1

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 $\frac {-5 + 12 \log 2} {18}$.

Answer 2

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