Question

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

• A. $lo{g}_e\left(\frac{4}{e}\right)$
• B. $2lo{g}_{e}2$
• C. $lo{g}_{e}2-1$
• D. $lo{g}_{e}2$

Answer 1

The correct option is A $lo{g}_e\left(\frac{4}{e}\right)$

$\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}\cdots \infty |T_n|=\frac{1}{n(n+1)}=(\frac{1}{n}-\frac{1}{n+1})S=T_1-T_2+T_3-T_4+T_5\cdots \infty =(\frac{1}{1}-\frac{1}{2})-(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})-(\frac{1}{4}-\frac{1}{4}-\frac{1}{5})1-2[\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\cdots \infty ]=1-2[-log(1+1)+1]=2log2-1=log\left(\frac{4}{e}\right).$