Question

$\left(\frac{1}{1.3}\right)+\left(\frac{1}{2.5}\right)+\left(\frac{1}{3.7}\right)+\left(\frac{1}{4.9}\right)+...is$

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

Answer 1

The correct option is B

$2-{\log }_{e}2$

Explanation for the correct option:

Step 1. Find the $nth$ term of the given series:

Let $S=\left(\frac{1}{1.3}\right)+\left(\frac{1}{2.5}\right)+\left(\frac{1}{3.7}\right)+\left(\frac{1}{4.9}\right)+...is$

Let $T_n$​ be the $nth$ term of the given series

$\Rightarrow T_n=\frac{1}{n(2n+1)}$ ……….(1)

Let $\frac{1}{n(2n+1)}=\frac{A}{n}+\frac{B}{2n+1}$

Step 2. Find the value of $A$ and $B$:
$\Rightarrow A=\underset{n\to 0}{\lim }n\left[\frac{1}{n(2n+1)}\right]$ ……(2)

$$\begin{gathered} =\underset{n\to 0}{\lim }\left[\frac{1}{2n+1}\right] \\ =\frac{1}{0+1} \\ =1 \end{gathered}$$

and $B=\underset{2n+1\to 0}{\lim }(2n+1)\left[\frac{1}{n(2n+1)}\right]$ …….(3)
$$\begin{gathered} =\underset{n\to \frac{-1}{2}}{\lim }\left[\frac{1}{n}\right] \\ =-2 \end{gathered}$$

Put the values of equation (2) and (3) in equation (1):
$\Rightarrow \frac{1}{n(2n+1)}=\frac{1}{n}-\frac{2}{2n+1}$
$\Rightarrow$ $Tn=\frac{1}{n}-\frac{2}{2n+1}$

Step 3. Find the sum of the series using expansion of ${\mathbf{log}}_{\mathbf{e}}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{)}$

$\begin{array}{rcl}S& =& \sum _{n=1}^{\infty }{T}_n\\ & =& \sum _{n=1}^{\infty }\left(\frac{1}{n}-\frac{2}{2n+1}\right)\end{array}$

$=\left(1-\frac{2}{3}\right)+\left(\frac{1}{2}-\frac{2}{5}\right)+\left(\frac{1}{3}-\frac{2}{7}\right)+...$

$=\left(1+\frac{1}{2}+\frac{1}{3}+...\right)-2\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...\right)$
$=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+...$
$=2-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+...\right)$

$=\mathbf{2}\mathbf{-}\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{e}}\mathbf{2}$ $\left[\mathbf{\because }{\mathbf{log}}_{\mathbf{e}}\mathbf{(}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathbf{x}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{x}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{4}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{)}\right]$

Hence, Option ‘B’ is Correct.