S-7

\(\displaystyle 7).\;\; \frac{1}{5}+\frac{1}{3}.\frac{1}{5^3}+\frac{1}{5}.\frac{1}{5^5}+\cdots\)

\(\text{We can also write it as}\)

\(\displaystyle \frac{1}{5}+\frac{\frac{1}{5^2}}{3}+\frac{\frac{1}{5^4}}{5}+\cdots \cdots \cdots\;\;\;\;(1)\)

\(\text{We have a result which we will use here to solve this expression}\)

\(\displaystyle \ln\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)\)

\(\text{So from above we have}\)

\(\displaystyle \frac{1}{2}.\ln\left(\frac{1+x}{1-x}\right)=\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)\)

\(\text{If you look above series and equation (1) you will find some similarity}\)

\(\text{Using above series we can find that for (1) x=1/5}\)

\(\displaystyle \left(\frac{1}{5}+\frac{\frac{1}{5^2}}{3}+\frac{\frac{1}{5^4}}{5}+\cdots\right) =\frac{1}{2}.\ln\left(\frac{1+\frac{1}{5}}{1-\frac{1}{5}}\right) \)

\(\text{So at the end we get}\)

\(\displaystyle \left(\frac{1}{5}+\frac{\frac{1}{5^2}}{3}+\frac{\frac{1}{5^4}}{5}+\cdots\right) =\frac{1}{2}.\ln\left(\frac{1+\frac{1}{5}}{1-\frac{1}{5}}\right) \)

\(\text{ANSWER :}\;\;\;\; \displaystyle \frac{1}{2}\ln\left(\frac{3}{2}\right)\)