Sum of series 4.

\(\displaystyle \;\; \sum_{n=2}^{\infty} \frac{(-1)^{n}}{n^{2}+n - 2}\)

SOLUTION

\(\displaystyle \Rightarrow \sum_{n=2}^{\infty} \frac{(-1)^{n}}{n^{2}+2n-n-2}\)

\(\displaystyle \Rightarrow \sum_{n=2}^{\infty} \frac{(-1)^n}{n(n+2)-(n+2)}\)

\(\displaystyle \Rightarrow \sum_{n=2}^{\infty} \frac{(-1)^{n}}{(n-1)(n+2)}\)

\(\displaystyle \Rightarrow\sum_{n=2}^{\infty}\left[\frac{(-1)^{n}}{-3(n-1)}+\frac{(-1)^{n}}{3(n+2)}\right]\)

\(\displaystyle \Rightarrow\frac{1}{3} \sum_{n=2}^{\infty}\left[\frac{(-1)^{n+1}}{n-1}+\frac{(-1)^{n}}{n+2}\right]\)

\(\displaystyle \Rightarrow \frac{1}{3} \sum_{n=2}^{\infty} \frac{(-1)^{n+1}}{n-1}-\frac{1}{3} \sum_{n=2}^{\infty} \frac{(-1)^{n-1}}{n+2}\;\; \cdots (1)\)

We have series of \(\log(1+x)\) as

\(\displaystyle \Rightarrow\log (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots\)

Using this in (1) , we will get

\(\displaystyle \Rightarrow-\frac{1}{3}[\log (2)]-\frac{1}{3}\left[\log 2-1+\frac{1}{2}-\frac{1}{3}\right]\)

\(\displaystyle \Rightarrow-\frac{2}{3} \log 2-\frac{1}{6}+\frac{1}{9}+\frac{1}{3}\)

\(\displaystyle \Rightarrow-\frac{2}{3} \log 2+\frac{-3+2+6}{18}\)

\(\displaystyle \Rightarrow-\frac{2}{3} \log 2+\frac{5}{18}\)

SOLUTION: \(\displaystyle \Rightarrow-\frac{2}{3} \log 2+\frac{5}{18}\)