Question

$\left( {\dfrac{1}{{1.3}}} \right) + \left( {\dfrac{1}{{2.5}}} \right) + \left( {\dfrac{1}{{3.7}}} \right) + \left( {\dfrac{1}{{4.9}}} \right) + .......$is
1. $2{\log _e}2 - 2$
2. $2 - {\log _e}2$
3. $2{\log _e}4$
4. ${\log _e}4$

Answer 1

Hint: Write the $nth$ term of the given series yourself. Break the term in two parts and find the value of constants $A$ and $B$ using limit $n \to 0$ for $A$ and $2n + 1 \to 0$ for $B$. Put the required values in the $nth$ term and find the sum from $n \to 1$ to $\infty $.

Formula Used:
${t_n} = \dfrac{1}{{n\left( {2n + 1} \right)}}$ (Sequence of given series)

Complete step by step Solution:
Let, ${t_n}$ be the $nth$ term of given series
${t_n} = \dfrac{1}{{n\left( {2n + 1} \right)}}$
${t_n} = \dfrac{A}{n} + \dfrac{B}{{2n + 1}}$
$ \Rightarrow \dfrac{1}{{n\left( {2n + 1} \right)}} = \dfrac{A}{n} + \dfrac{B}{{2n + 1}} - - - - - (1)$
$\therefore A = \mathop {\lim }\limits_{n \to 0} n\left( {\dfrac{1}{{n\left( {2n + 1} \right)}}} \right),B = \mathop {\lim }\limits_{2n + 1 \to 0} \left( {2n + 1} \right)\left( {\dfrac{1}{{n\left( {2n + 1} \right)}}} \right)$
$A = \mathop {\lim }\limits_{n \to 0} \left( {\dfrac{1}{{\left( {2n + 1} \right)}}} \right),B = \mathop {\lim }\limits_{n \to \dfrac{{ - 1}}{2}} \left( {\dfrac{1}{n}} \right)$
$A = 1,B = - 2$
Put $A = 1,B = - 2$ in equation (1)
$\dfrac{1}{{n\left( {2n + 1} \right)}} = \dfrac{1}{n} - \dfrac{2}{{2n + 1}}$ OR ${t_n} = \dfrac{1}{n} - \dfrac{2}{{2n + 1}}$
Now the sum of the series $\left( {\dfrac{1}{{1.3}}} \right) + \left( {\dfrac{1}{{2.5}}} \right) + \left( {\dfrac{1}{{3.7}}} \right) + \left( {\dfrac{1}{{4.9}}} \right) + .......$ is,
$S = \sum\limits_{n = 1}^\infty {{t_n}} $
$ = \sum\limits_{n = 1}^\infty {\dfrac{1}{n} - \dfrac{2}{{2n + 1}}} $
$ = \left( {1 - \dfrac{2}{3}} \right) + \left( {\dfrac{1}{2} - \dfrac{2}{5}} \right) + \left( {\dfrac{1}{3} - \dfrac{2}{7}} \right) + ......$
$ = \left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + ......} \right) - 2\left( {\dfrac{1}{3} + \dfrac{1}{5} + \dfrac{1}{7} + ......} \right)$
$ = 1 + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{1}{5} + \dfrac{1}{6} - \dfrac{1}{7} + ......$
$ = 2 - 2 + 1 + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{1}{5} + \dfrac{1}{6} - \dfrac{1}{7} + ......$
$ = 2 - 1 + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{1}{5} + \dfrac{1}{6} - \dfrac{1}{7} + ......$
$ = 2 - \left( {1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - \dfrac{1}{6} + \dfrac{1}{7} - ......} \right)$
$ = 2 - \log {}_e2$

Hence, the correct option is 2.

Note: The key concept involved in solving this problem is a good knowledge of Series and Sequence. Students must remember that a finite series is formed by adding all the terms of a finite sequence and an infinite series is formed by adding all the terms in an infinite sequence.