Question

The sum of the series $[1+(\frac{1}{2}+\frac{1}{3}).\frac{1}{4}+(\frac{1}{4}+\frac{1}{5}).\frac{1}{4^2}+(\frac{1}{6}+\frac{1}{7})\frac{1}{4^3}+....\infty ]$ is equal to

• A. $log\sqrt{12}$
• B. $log\sqrt{10}$
• C. $e^{12}$
• D. None of these

Answer 1

The correct option is A $log\sqrt{12}$

$S=[1+(\frac{1}{2}+\frac{1}{3})\frac{1}{4}+(\frac{1}{4}+\frac{1}{5})\frac{1}{4^2}+(\frac{1}{6}+\frac{1}{7})\frac{1}{4^3}+\cdot \cdot \cdot \infty ]$

$S=S_1+S_2$

$S_1=[1+\left(\frac{1}{3}\right)\frac{1}{4}+\left(\frac{1}{5}\right)\frac{1}{4^2}+\left(\frac{1}{7}\right)\frac{1}{4^3}+\cdot \cdot \cdot \infty ]$

$S_1={\sum }_{n=0}^{\infty }\left(\frac{1}{2n+1}\right)\frac{1}{4^n}$

$\frac{1}{2n+1}={\int }_0^1{x}^{2n}dx$

$S_1={\sum }_{n=0}^{\infty }\left({\int }_0^1{x}^{2n}dx\right)\frac{1}{4^n}$

$S_1={\int }_0^1{\sum }_{n=0}^{\infty }{\left(\frac{x^2}{4}\right)}^{n}dx$

$S_1={\int }_0^1\frac{1}{1-\frac{x^2}{4}}dx$

Thus $S_1=\ln 3$

Similarly , $S_2=[\left(\frac{1}{2}\right)\frac{1}{4}+\left(\frac{1}{4}\right)\frac{1}{4^2}+\left(\frac{1}{6}\right)\frac{1}{4^3}+\cdot \cdot \cdot \infty ]$

$S_2={\sum }_{n=1}^{\infty }\left(\frac{1}{2n}\right)\frac{1}{4^n}$

$\frac{1}{2n}={\int }_0^1{x}^{2n-1}dx$

$S_2={\sum }_{n=1}^{\infty }\left({\int }_0^1{x}^{2n-1}dx\right)\frac{1}{4^n}$

$S_2={\int }_0^1{\sum }_{n=1}^{\infty }{\left(\frac{x^2}{4}\right)}^n\frac{1}{x}dx$

$S_2={\int }_0^1\frac{\frac{x}{4}}{1-\frac{x^2}{4}}dx$

$S_2=-\frac{1}{2}\ln \frac{3}{4}dx$

$S=S_1+S_2$

$S=\ln 3-\frac{1}{2}\ln \frac{3}{4}$

$S=\ln \sqrt{12}=\log \sqrt{12}$