Solveeit Logo
Login

Question

Question: The sum of series \(\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{5.6.7}+....\) is (1) \({{\log }_{e...

The sum of series 11.2.3+13.4.5+15.6.7+....\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{5.6.7}+.... is
(1) log⁡e2−12{{\log }_{e}}2-\dfrac{1}{2}
(2) log⁡e2{{\log }_{e}}2
(3) log⁡e2+12{{\log }_{e}}2+\dfrac{1}{2}
(4) log⁡e2+1{{\log }_{e}}2+1

Explanation

Solution

Here in this question we have been asked to find the sum of series 11.2.3+13.4.5+15.6.7+....\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 1(2n−1)(2n)(2n+1)\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 11.2.3+13.4.5+15.6.7+....\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 1(2n−1)(2n)(2n+1)\dfrac{1}{\left( 2n-1 \right)\left( 2n \right)\left( 2n+1 \right)} .
Let us assume that 1(2n−1)(2n)(2n+1)=(12)[A2n−1+Bn+C2n+1]\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(2n+1)n+B(2n+1)(2n−1)+Cn(2n−1)=1A\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=1A=1 , B=−1B=-1 and C=1C=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 SS .
Now we can say that 2S=(11−11+13)+(13−12+15)+(15−13+17)+.......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=(−12+13−14+15−16+27+.......)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(1−12+13−14+15−16+..............)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⁡ex=∑n=1∞(−1)n−1(x−1)nn{{\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⁡e2=1−12+13−14+15−16+..............{{\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=2log⁡e21+2S=2{{\log }_{e}}2 then we will have S=log⁡e2−12S={{\log }_{e}}2-\dfrac{1}{2} .
Therefore we can conclude that the sum of the given series 11.2.3+13.4.5+15.6.7+....\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{5.6.7}+.... will be given as log⁡e2−12{{\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(x+1)=∑n=1∞(−1)n−1(x)nn{{\log }_{e}}\left( x+1 \right)=\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( -1 \right)}^{n-1}}{{\left( x \right)}^{n}}}{n}} .