Question

The sum of the series $\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-...$ upto $\infty$ is equal to

• A. ${\log }_{e}2-1$
• B. ${\log }_{e}2$
• C. ${\log }_e\frac{4}{e}$
• D. $2{\log }_{e}2$

Answer 1

Answer: (C)

${\log }_e\frac{4}{e}$

$T_n=(-1{)}^{n+1}\frac{1}{n(n+1)}$
$=(-1{)}^{n+1}\frac{n+1-n}{n(n+1)}$
$=(-1{)}^{n+1}[\frac{1}{n}-\frac{1}{n+1}]$
$S=1-\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...\infty$
We know that
$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}...\infty$
$S=[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}...\infty ]+[-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}...\infty ]$
$=[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}...\infty ]+[-1+1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}...\infty ]$
$={\log }_e\left(2\right)+{\log }_e\left(2\right)-1$
$={\log }_e\left(4\right)-{\log }_e\left(e\right)$
$S={\log }_e\left(\frac{4}{e}\right)$