Question

Find the sum of the series $\dfrac{1}{2\cdot 3}+\dfrac{1}{4\cdot 5}+\dfrac{1}{6\cdot 7}+...\infty =$
A. ${{\log }_{e}}\left( \dfrac{2}{e} \right)$
B. ${{\log }_{e}}\left( \dfrac{e}{2} \right)$
C. $\dfrac{2}{e}$
D. $\dfrac{e}{2}$

Answer 1

Hint: In this question, we are to find the sum of the given series. For this, we need to apply the logarithmic series formula. By rewriting the series into the form of a logarithmic series, we can evaluate the sum of the given series.



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}{2\cdot 3}+\dfrac{1}{4\cdot 5}+\dfrac{1}{6\cdot 7}+...\infty $
Rewriting the given series as
$\begin{align}
  & \dfrac{1}{2\cdot 3}+\dfrac{1}{4\cdot 5}+\dfrac{1}{6\cdot 7}+...\infty =\left( \dfrac{1}{2}-\dfrac{1}{3} \right)+\left( \dfrac{1}{4}-\dfrac{1}{5} \right)+\left( \dfrac{1}{6}-\dfrac{1}{7} \right)+...\infty \\
 & \Rightarrow \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}+....\infty \\
\end{align}$
But we a logarithmic series as
$\begin{align}
  & {{\log }_{e}}(1+x)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+... \\
 & \Rightarrow x-{{\log }_{e}}(1+x)=\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{4}}}{4}+... \\
\end{align}$
On comparing (1) and (2), we get
$x=1$
Thus, on substituting this value into the logarithmic function, we get
$\begin{align}
  & \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}+....\infty =1-{{\log }_{e}}(1+x) \\
 & \Rightarrow 1-{{\log }_{e}}(1+1) \\
 & \Rightarrow 1-{{\log }_{e}}2 \\
\end{align}$
Since we know that the base of the log and the argument of the log are the same, then its value is equal to $1$, on applying this property, we get
${{\log }_{e}}e=1$
Then,
$1-{{\log }_{e}}2={{\log }_{e}}e-\log 2$
We have the property of logarithms,
$\log x-\log y=\log \left( \dfrac{x}{y} \right)$
We can write
$\Rightarrow \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}+....\infty ={{\log }_{e}}e-\log 2={{\log }_{e}}\left( \dfrac{e}{2} \right)$
Thus, the sum is
$\dfrac{1}{5}+\dfrac{1}{2}\cdot \dfrac{1}{{{5}^{2}}}+\dfrac{1}{3}\cdot \dfrac{1}{{{5}^{3}}}+...\infty ={{\log }_{e}}\left( \dfrac{e}{2} \right)$



Option ‘B’ is correct



Note: In this question, the series is easy to compare, the only difference is the variable $x$. By using logarithmic functions and expansions, this type of sum’s would be evaluated. We have to apply logarithm properties if needed in the simplification. Here, on comparing we know that the series is same type but the variable is different i.e., in the given series $x=1$. So, by substituting this in the logarithmic function, the required sum will be obtained.