Question

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

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

Answer 1

The correct option is C $2{\log }_{e}2-1$

$\frac{1}{1\times 2}-\frac{1}{2\times 3}+\frac{1}{3+4}-\frac{1}{4\times 5}+\cdots$

$(1-\frac{1}{2})-(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})-(\frac{1}{4}-\frac{1}{5})+\cdots$

$(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\cdots )+(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\cdots )$

$\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$

Put $x=1$

${\log }_2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots \cdot$

And

$-1+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\cdots \cdot =\log 2-1$

Hence

$=\left({\log }_2\right)+({\log }_2-1)$

$=2\log 2-1$