Question

$1+(\frac{1}{2}+\frac{1}{3})\frac{1}{4}+(\frac{1}{4}+\frac{1}{5})\frac{1}{4^2}+(\frac{1}{6}+\frac{1}{7})\frac{1}{4^3}+...\infty ={\log }_e\sqrt{b}$.
Find $\frac{b}{4}$

Answer 1

$1+(\frac{1}{2}+\frac{1}{3})\frac{1}{4}+(\frac{1}{4}+\frac{1}{5})\frac{1}{4^2}+(\frac{1}{6}+\frac{1}{7})\frac{1}{4^3}+...\infty$

$=(\frac{1}{2}.\frac{1}{4}+\frac{1}{4}.\frac{1}{4^2}+\frac{1}{6}.\frac{1}{4^3}+...)+(1+\frac{1}{3}.\frac{1}{4}+\frac{1}{5}.\frac{1}{4^2}+\frac{1}{7}.\frac{1}{4^3}+...)$

$=\frac{1}{2}[\frac{1}{4}+\frac{1}{2}{\left(\frac{1}{4}\right)}^2+\frac{1}{3}{\left(\frac{1}{4}\right)}^3+...]+2[\frac{1}{2}+\frac{1}{3}{\left(\frac{1}{2}\right)}^3+\frac{1}{5}{\left(\frac{1}{2}\right)}^5+...]$

$=\frac{1}{2}(-{\log }_e(1-\frac{1}{4}))+{\log }_e\left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right)$

$=-\frac{1}{2}{\log }_e\left(\frac{3}{4}\right)+{\log }_e\left(3\right)$

$={\log }_e\left(3\right)-{\log }_e\sqrt{\frac{3}{4}}$

$={\log }_e\left(3\right)+{\log }_e\frac{2}{\sqrt{3}}$

$={\log }_e\left(3.\frac{2}{\sqrt{3}}\right)$

$={\log }_{e}2\sqrt{3}$

$={\log }_e\sqrt{12}$

$\Rightarrow \frac{b}{4}=3$