Question

$\sum _{n=1}^{\infty }\frac{1}{2n(2n+1)}$ is equal to

• A. ${\log }_e\left(\frac{2}{e}\right)$
• B. $1-{\log }_{e}2$
• C. $1-{\log }_e\frac{1}{2e}$
• D. none of these

Answer 1

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

Given, $\frac{1}{2n(2n+1)}$
$=\frac{2n+1-2n}{2n+1\left(2n\right)}$
$=\frac{1}{2n}-\frac{1}{2n+1}$
Therefore applying, summation we get
$\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}...\infty$
Let $\alpha =\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}...\infty$
$\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}...\infty$
Substituting x=1, we get
$\log \left(2\right)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}...\infty$
$\log \left(2\right)=1-\alpha$ ...(from i)
Hence
$\alpha =1-{\log }_e\left(2\right)$