Question

Find $\dfrac{1}{1\cdot 2\cdot 3}+\dfrac{1}{3.4.5}+\dfrac{1}{5\cdot 6\cdot 7}+.....\infty =$
A. ${{\log }_{e}}\sqrt{2}$
B. ${{\log }_{e}}2-\dfrac{1}{2}$
C. ${{\log }_{e}}2$
D. ${{\log }_{e}}4$

Answer 1

Hint: In this question, we are to find the sum of the given series. This is done by using logarithmic series. By substituting $x=1$ in the logarithmic series, we get a series of terms with that we can frame the required series. Then, on simplifying, we get the required sum.



Formula Used:Logarithmic series:
If $x\in R$ and $\left| x \right|<1$, then the expansion is
${{\log }_{e}}(1+x)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+...$
If $x\in R$ and $\left| x \right|<1$, then the expansion is
${{\log }_{e}}(1-x)=-x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-...$
Some important formulae:
If $\left| x \right|<1$ then ${{\log }_{e}}\dfrac{1+x}{1-x}=2\left( x+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}+... \right)$
If $x>1$ then ${{\log }_{e}}\dfrac{x+1}{x-1}=2\left( \dfrac{1}{x}+\dfrac{1}{3{{x}^{3}}}+\dfrac{1}{5{{x}^{5}}}+... \right)$



Complete step by step solution:Given series is
$\dfrac{1}{1\cdot 2\cdot 3}+\dfrac{1}{3.4.5}+\dfrac{1}{5\cdot 6\cdot 7}+.....\infty $
To solve this, we are using the logarithmic series.
The logarithmic series is written as
${{\log }_{e}}(1+x)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+...$
On substituting $x=1$, we get
$\begin{align}
  & {{\log }_{e}}(1+1)=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+.....\infty \text{ }...(1) \\
 & \Rightarrow {{\log }_{e}}2=\left( 1-\dfrac{1}{2} \right)+\left( \dfrac{1}{3}-\dfrac{1}{4} \right)+\left( \dfrac{1}{5}-\dfrac{1}{6} \right)+.....\infty \\
 & \Rightarrow {{\log }_{e}}2=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+.....\infty \text{ }...(2) \\
\end{align}$
Equation (1) is also written as
\[\begin{align}
  & {{\log }_{e}}(1+1)=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+.....\infty \\
 & \Rightarrow {{\log }_{e}}2=1-\left( \dfrac{1}{2}-\dfrac{1}{3} \right)-\left( \dfrac{1}{4}-\dfrac{1}{5} \right)-.....\infty \\
 & \Rightarrow {{\log }_{e}}2=1-\dfrac{1}{2\cdot 3}-\dfrac{1}{4\cdot 5}-.....\infty \text{ }....(3) \\
\end{align}\]
On adding (2) and (3), we get
$\begin{align}
  & {{\log }_{e}}2+{{\log }_{e}}2=\left( \dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+.....\infty \right)+\left( 1-\dfrac{1}{2\cdot 3}-\dfrac{1}{4\cdot 5}-.....\infty \text{ } \right) \\
 & \Rightarrow 2{{\log }_{e}}2=1+\left( \dfrac{1}{1.2}-\dfrac{1}{2\cdot 3} \right)+\left( \dfrac{1}{3.4}-\dfrac{1}{4\cdot 5} \right)+....\infty \\
 & \Rightarrow 2{{\log }_{e}}2-1=\left( \dfrac{2}{1\cdot 2\cdot 3} \right)+\left( \dfrac{2}{3\cdot 4\cdot 5} \right)+....\infty \\
 & \Rightarrow \dfrac{1}{2}\left[ 2{{\log }_{e}}2-1 \right]=\left( \dfrac{1}{1\cdot 2\cdot 3} \right)+\left( \dfrac{1}{3\cdot 4\cdot 5} \right)+....\infty \text{ }...(4) \\
\end{align}$
Equation (4) represents the given series.
Thus, we can write
\[\begin{align}
  & \dfrac{1}{1\cdot 2\cdot 3}+\dfrac{1}{3.4.5}+\dfrac{1}{5\cdot 6\cdot 7}+.....\infty =\dfrac{1}{2}\left[ 2{{\log }_{e}}-1 \right] \\
 & \Rightarrow {{\log }_{e}}2-\dfrac{1}{2} \\
\end{align}\]
Therefore, the sum is ${{\log }_{e}}2-\dfrac{1}{2}$.



Option ‘B’ is correct



Note: Here, we need to apply the logarithmic series formula, where we need to submit $x=1$. So, that we are able to simplify the terms and frame them in such a way that we get the given series. In this way, this type of question is to be solved. Here we need to remember that, the subtraction of two rational numbers will result in the product of their denominators. This is just a simple logic that taking L.C.M. But it is written in the product form. This is because the given question is in such a way.