Question

The sum of series $\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{5.6.7}+....$ is
(1) ${{\log }_{e}}2-\dfrac{1}{2}$
(2) ${{\log }_{e}}2$
(3) ${{\log }_{e}}2+\dfrac{1}{2}$
(4) ${{\log }_{e}}2+1$

Answer 1

Hint: Here in this question we have been asked to find the sum of series $\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{5.6.7}+....$. To answer this question, first we will evaluate the general term of the given series that is $\dfrac{1}{\left( 2n-1 \right)\left( 2n \right)\left( 2n+1 \right)}$ . Now we will evaluate the summation of the series.

Complete step-by-step solution:
Now considering from the question we have been asked to find the sum of series $\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{5.6.7}+....$ .
To answer this question, first we will evaluate the general term of the given series that is $\dfrac{1}{\left( 2n-1 \right)\left( 2n \right)\left( 2n+1 \right)}$ .
Let us assume that $\dfrac{1}{\left( 2n-1 \right)\left( 2n \right)\left( 2n+1 \right)}=\left( \dfrac{1}{2} \right)\left[ \dfrac{A}{2n-1}+\dfrac{B}{n}+\dfrac{C}{2n+1} \right]$ from this we will have $A\left( 2n+1 \right)n+B\left( 2n+1 \right)\left( 2n-1 \right)+Cn\left( 2n-1 \right)=1$ by simplifying this equation we will have $A=1$ , $B=-1$ and $C=1$ .
Now we will have $\dfrac{1}{2}\left\{ \dfrac{1}{2n-1}-\dfrac{1}{n}+\dfrac{1}{2n+1} \right\}$ .
Let us assume that the summation of the series is $S$ .
Now we can say that $2S=\left( \dfrac{1}{1}-\dfrac{1}{1}+\dfrac{1}{3} \right)+\left( \dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{5} \right)+\left( \dfrac{1}{5}-\dfrac{1}{3}+\dfrac{1}{7} \right)+.......$ .
Here we will have $2S=\left( -\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{2}{7}+....... \right)$ .
By further simplifying this expression we will have $1+2S=2\left( 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+.............. \right)$ .
From the basic concepts, we know that ${{\log }_{e}}x=\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( -1 \right)}^{n-1}}{{\left( x-1 \right)}^{n}}}{n}}$ .
By using this we will have ${{\log }_{e}}2=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+..............$ .
Hence we can say that $1+2S=2{{\log }_{e}}2$ then we will have $S={{\log }_{e}}2-\dfrac{1}{2}$ .
Therefore we can conclude that the sum of the given series $\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{5.6.7}+....$ will be given as ${{\log }_{e}}2-\dfrac{1}{2}$.
Hence we will mark the option “1” as correct.

Note: During the process of answering questions of this type we should be sure with the calculations that we are going to perform in between the steps in order to simplify it further. From the basic concepts, we know that the formula for the expansion of logarithm is given as ${{\log }_{e}}\left( x+1 \right)=\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( -1 \right)}^{n-1}}{{\left( x \right)}^{n}}}{n}}$ .