CSIR JUNE 2011 PART B QUESTION 32 SOLUTION (Find $\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n(n+1)}.$)

Using the fact that $\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} =log 2$, find the value of $$\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n(n+1)}.$$
Solution: We have $\frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1}$, $$\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n(n+1)} = \sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n} - \sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n+1} = (-1)\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} - \sum\limits_{n=1}^{\infty}\frac{(-1)^{n+2}}{n+1}.$$ This is equal to $$-log 2 - (\sum\limits_{n=2}^{\infty}\frac{(-1)^{n+1}}{n})) = -log 2 - ((\sum\limits_{n=2}^{\infty}\frac{(-1)^{n+1}}{n}-\frac{(-1)^1}{1})+\frac{(-1)^1}{1}) = 1-2log 2.$$

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