Question

The sum of the series $\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+....+\infty$ =

• A. $log(2e)$
• B. $log(e/2)$
• C. $log(4/e)$
• D. $Noneofthese$

Answer 1

Answer: (B)

$\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\dots +\infty$
Here, $T_n=\frac{1}{2n.\left(2n+1\right)}$
$T_n=\frac{1}{2n}-\frac{1}{2n+1}$
$T_n=\frac{1}{2n}-\frac{1}{2n+1}$
$T_1=\frac{1}{2}-\frac{1}{3}$
$T_2=\frac{1}{4}-\frac{1}{5}$
$T_3=\frac{1}{6}-\frac{1}{7}\dots \infty$
$\dots \dots \dots \dots \dots \dots \dots \dots \dots$
$\dots \dots \dots \dots \dots \dots \dots \dots \dots$
Adding all, $\left(T_1+T_2+T_3+\dots +T_{\infty }\right)$
$=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\dots \right)-\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots \right)\dots \left(i\right)$
We know that, $log2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots \infty$
From Eq. $\left(i\right)$,
$\left(T_1+T_2+\dots +T_{\infty }\right)=-\left(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots \infty \right)$
$=-\left\{\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots \infty \right)-1\right\}$
$=-\left\{lo{g}_{e}2-lo{g}_{e}e\right\}$
$=-log\left(2/e\right)$
$=log\left(e/2\right)$