S-4

\(\displaystyle 4).\;\; \frac{1}{3}+\frac{1}{4}.\frac{1}{2!} +\frac{1}{5}.\frac{1}{3!}\)

\(\text{Above expresion can be written in form}\)

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

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

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

\(\text{We use}\;\;e^x\;\; \text{Series expension and derived a result}\)

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

\(\text{So the (1) will become}\)

\(\displaystyle e-2 -\left (e-2-\frac{1}{2}\right)\)

\(\text{ANSWER :}\;\; \displaystyle \frac{1}{2}\)