S-10

\(\displaystyle 10).\;\; \sum_{n=1}^{\infty} \frac{(-1)^n}{n(n+1)}\)

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

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

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

\(\text{We have a series expension , we will use that in (1)}\)

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

\(\text{So by using above in (1) we will get }\)

\(\displaystyle \Rightarrow -\ln 2 -(\ln 2)-1\)

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