S-5

\(\displaystyle 5).\;\; \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n+1)}\)

\(\text{Above Expression can be written as }\)

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

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

\(\text{We know the series expension of }\)

\(\displaystyle\ln(2) = 1-\frac{1}{2}+\frac{1}{3} -\frac{1}{4} +\cdots\)

\(\text{Using above we can write}\)

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

\(\text{Form above we can write }\;\; (1) \;\; \text{as}\)

\(\ln(2) +\ln(2)-1\)

\(\text{ANSWER :}\;\;\;\; 2\ln(2)-1\)