Question

Sum of the series \[\dfrac{1}{1.2.3}+\dfrac{5}{3.4.5}+\dfrac{9}{5.6.7}+...\] is equal to
A. $ \dfrac{3}{2}-3{{\log }_{e}}2 $
B. $ \dfrac{5}{2}-3{{\log }_{e}}2 $
C. $ 1-4{{\log }_{e}}2 $
D. none of these

Answer 1

Hint: We first try to find the general form of the given series \[\dfrac{1}{1.2.3}+\dfrac{5}{3.4.5}+\dfrac{9}{5.6.7}+...\]. We take A.P. series in the numerator and the G.P. series in the denominator. We break the general form using the factors of the denominator. We find the diverging sequence and solve it.

Complete step-by-step answer:
We first need to find the general term of the series \[\dfrac{1}{1.2.3}+\dfrac{5}{3.4.5}+\dfrac{9}{5.6.7}+...\].
The numerator of the terms is an A.P. series and the mid terms of the denominator are G.P. series.
We find the general form of 1, 5, 9, ……. Which gives a common difference as $ 9-5=4 $ .
Therefore, the $ {{n}^{th}} $ term will be $ 1+4\left( n-1 \right)=4n-3 $ .
The mid terms of the denominators are 2, 4, 6, ……. Which gives the general form as $ 2n $ .
The other terms are $ 2n-1 $ and $ 2n+1 $ .
Therefore, general $ {{n}^{th}} $ term of the series \[\dfrac{1}{1.2.3}+\dfrac{5}{3.4.5}+\dfrac{9}{5.6.7}+...\] is \[\dfrac{4n-3}{\left( 2n-1 \right)2n\left( 2n+1 \right)}\].
We want to break the numerator with respect to denominator.
So, we get \[4n-3=2\left( 2n-1 \right)+2n-\left( 2n+1 \right)\].
So, \[\dfrac{4n-3}{\left( 2n-1 \right)2n\left( 2n+1 \right)}=\dfrac{2\left( 2n-1 \right)+2n-\left( 2n+1 \right)}{\left( 2n-1 \right)2n\left( 2n+1 \right)}=\dfrac{1}{n\left( 2n+1 \right)}+\dfrac{1}{\left( 2n-1 \right)\left( 2n+1 \right)}-\dfrac{1}{\left( 2n-1 \right)2n}\]
We again break the series.
\[\dfrac{1}{n\left( 2n+1 \right)}=\dfrac{2n+1-2n}{n\left( 2n+1 \right)}=\dfrac{1}{n}-\dfrac{2}{\left( 2n+1 \right)}\]
\[\dfrac{1}{\left( 2n-1 \right)\left( 2n+1 \right)}=\dfrac{1}{2}\times \dfrac{\left( 2n+1 \right)-\left( 2n-1 \right)}{\left( 2n-1 \right)\left( 2n+1 \right)}=\dfrac{1}{2\left( 2n-1 \right)}-\dfrac{1}{2\left( 2n+1 \right)}\]
\[-\dfrac{1}{\left( 2n-1 \right)2n}=\dfrac{\left( 2n-1 \right)-2n}{\left( 2n-1 \right)2n}=\dfrac{1}{2n}-\dfrac{1}{\left( 2n-1 \right)}\]
We get
\[\begin{align}
  & \dfrac{4n-3}{\left( 2n-1 \right)2n\left( 2n+1 \right)} \\
 & =\dfrac{1}{n}-\dfrac{2}{\left( 2n+1 \right)}+\dfrac{1}{2\left( 2n-1 \right)}-\dfrac{1}{2\left( 2n+1 \right)}+\dfrac{1}{2n}-\dfrac{1}{\left( 2n-1 \right)} \\
 & =\dfrac{3}{2n}-\dfrac{5}{2\left( 2n+1 \right)}-\dfrac{1}{2\left( 2n-1 \right)} \\
\end{align}\]
So, \[\dfrac{1}{1.2.3}+\dfrac{5}{3.4.5}+\dfrac{9}{5.6.7}+...=\sum\limits_{n-1}^{\infty }{\dfrac{3}{2n}}-\sum\limits_{n-1}^{\infty }{\dfrac{5}{2\left( 2n+1 \right)}}-\sum\limits_{n-1}^{\infty }{\dfrac{1}{2\left( 2n-1 \right)}}\]
We know that $ {{\log }_{e}}\left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-....+{{\left( -1 \right)}^{n-1}}\dfrac{{{x}^{n}}}{n}+....\infty $ .
We also have
\[\begin{align}
  & \sum\limits_{n-1}^{\infty }{\dfrac{3}{2n}}=\dfrac{3}{2}\left[ 1+\dfrac{1}{2}+\dfrac{1}{3}+....\infty \right] \\
 & \sum\limits_{n-1}^{\infty }{\dfrac{5}{2\left( 2n+1 \right)}}=\dfrac{5}{2}\left[ \dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+....\infty \right] \\
 & \sum\limits_{n-1}^{\infty }{\dfrac{1}{2\left( 2n-1 \right)}}=\dfrac{1}{2}\left[ 1+\dfrac{1}{3}+\dfrac{1}{5}+....\infty \right] \\
\end{align}\]
The sum form of \[1+\dfrac{1}{2}+\dfrac{1}{3}+....\infty \] is not a converging series. So, \[1+\dfrac{1}{2}+\dfrac{1}{3}+....\infty \to \infty \].
\[\dfrac{1}{1.2.3}+\dfrac{5}{3.4.5}+\dfrac{9}{5.6.7}+...=\sum\limits_{n-1}^{\infty }{\dfrac{3}{2n}}-\sum\limits_{n-1}^{\infty }{\dfrac{5}{2\left( 2n+1 \right)}}-\sum\limits_{n-1}^{\infty }{\dfrac{1}{2\left( 2n-1 \right)}}\to \infty \].
The correct option is D.
So, the correct answer is “Option D”.

Note: The series becomes diverging as the $ {{\log }_{e}}\left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-....+{{\left( -1 \right)}^{n-1}}\dfrac{{{x}^{n}}}{n}+....\infty $ formula changes to $ {{\log }_{e}}\left( 1-x \right)=-\left[ x+\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+....+\dfrac{{{x}^{n}}}{n}+....\infty \right] $ . The value we are putting is 1 in place of $ x $ . Therefore, we can’t have the value of $ {{\log }_{e}}0 $ .