Question

The sum of the infinite series $\frac{1}{2}\left(\frac{1}{3}+\frac{1}{4}\right)-\frac{1}{4}\left(\frac{1}{3^{2}}+\frac{1}{4^{2}}\right)+\frac{1}{6}\left(\frac{1}{3^{3}}+\frac{1}{4^{3}}\right)-...$ is equal to

• A. $\frac{1}{2} log\,2$
• B. $log \frac{3}{5}$
• C. $log \frac{5}{3}$
• D. $\frac{1}{2} log \frac{5}{3}$

Answer 1

Answer: (D)

Consider
$\frac{1}{2}\left(\frac{1}{3}+\frac{1}{4}\right)-\frac{1}{4}\left(\frac{1}{3^{2}}+\frac{1}{4^{2}}\right)+\frac{1}{6}\left(\frac{1}{3^{3}}+\frac{1}{4^{3}}\right)-...$
$=\left(\frac{1}{2}. \frac{1}{3}-\frac{1}{4}.\frac{1}{3^{2}}+\frac{1}{6}. \frac{1}{3^{3}}...\right)+\left(\frac{1}{2}. \frac{1}{4}-\frac{1}{4}. \frac{1}{4^{2}}+\frac{1}{6}. \frac{1}{4^{3}}-\right)...$
$=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2}\left(\frac{1}{3^{2}}\right)+\frac{1}{3}\left(\frac{1}{3^{3}}\right)...\right)+\frac{1}{2}\left(\frac{1}{4}-\frac{1}{2}. \frac{1}{4^{2}}+\frac{1}{3}. \frac{1}{4^{3}}..\right)$
$=\frac{1}{2}\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}...\right)+\frac{1}{2}\left(y-\frac{y^{2}}{2}+\frac{Y^{3}}{3}\right)$
where $x=\frac{1}{3}, y=\frac{1}{4}$
$=\frac{1}{2}log\left(1+x\right)+\frac{1}{2}log\left(1+y\right)$
$=\frac{1}{2}log\left(1+\frac{1}{3}\right)+\frac{1}{2} log\left(1+\frac{1}{4}\right)=\frac{1}{2}log \frac{5}{3}$