Question

Find $\dfrac{x-1}{(x+1)}+\dfrac{1}{2}\cdot \dfrac{{{x}^{2}}-1}{{{(x+1)}^{2}}}+\dfrac{1}{3}\cdot \dfrac{{{x}^{3}}-1}{{{(x+1)}^{3}}}+.....\infty $
A. \[{{\log }_{e}}x\]
B. \[{{\log }_{e}}(1+x)\]
C. \[{{\log }_{e}}(1-x)\]
D. \[{{\log }_{e}}\dfrac{x}{1+x}\]

Answer 1

Hint: In this question, we are to find the sum of the given series. To solve this, the given series is to be rewritten in such a way that we can frame the series into a particular progression. So, by applying the appropriate formula, the required sum is to be calculated.



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{x-1}{(x+1)}+\dfrac{1}{2}\cdot \dfrac{{{x}^{2}}-1}{{{(x+1)}^{2}}}+\dfrac{1}{3}\cdot \dfrac{{{x}^{3}}-1}{{{(x+1)}^{3}}}+.....\infty $
Rewriting the given series as
\[\begin{align}
  & \Rightarrow \left( \dfrac{x}{x+1}-\dfrac{1}{x+1} \right)+\dfrac{1}{2}\left( \dfrac{{{x}^{2}}}{{{(x+1)}^{2}}}-\dfrac{1}{{{(x+1)}^{2}}} \right)+\dfrac{1}{3}\left( \dfrac{{{x}^{3}}}{{{(x+1)}^{3}}}-\dfrac{1}{{{(x+1)}^{3}}} \right)+.....\infty \\
 & \Rightarrow \left( \dfrac{x}{x+1}+\dfrac{1}{2}{{\left( \dfrac{x}{x+1} \right)}^{2}}+\dfrac{1}{3}{{\left( \dfrac{x}{x+1} \right)}^{3}}+....\infty \right)-\left( \dfrac{1}{x+1}+\dfrac{1}{2\cdot {{(x+1)}^{2}}}+\dfrac{1}{3\cdot {{(x+1)}^{3}}}+...\infty \right) \\
\end{align}\]
Then, we can write it as
\[\Rightarrow \left( \dfrac{x}{x+1}+\dfrac{{{\left( \dfrac{x}{x+1} \right)}^{2}}}{2}+\dfrac{{{\left( \dfrac{x}{x+1} \right)}^{3}}}{3}+...\infty \right)-\left( \dfrac{1}{x+1}+\dfrac{{{\left( \dfrac{1}{x+1} \right)}^{2}}}{2}+\dfrac{{{\left( \dfrac{1}{x+1} \right)}^{2}}}{3}+...\infty \right)\]
These are the logarithmic sequences. So, we can use the formula:
$\begin{align}
  & {{\log }_{e}}(1-x)=-x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-...\infty \\
 & \Rightarrow -{{\log }_{e}}(1-x)=x+\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+....\infty \\
\end{align}$
So, on applying this formula to the above two sequences, we get
\[\Rightarrow \left( \dfrac{x}{x+1}+\dfrac{{{\left( \dfrac{x}{x+1} \right)}^{2}}}{2}+\dfrac{{{\left( \dfrac{x}{x+1} \right)}^{3}}}{3}+...\infty \right)=-{{\log }_{e}}\left( 1-\dfrac{x}{x+1} \right)\]
$\Rightarrow \left( \dfrac{1}{x+1}+\dfrac{{{\left( \dfrac{1}{x+1} \right)}^{2}}}{2}+\dfrac{{{\left( \dfrac{1}{x+1} \right)}^{2}}}{3}+...\infty \right)=-{{\log }_{e}}\left( 1-\dfrac{1}{x+1} \right)$
Then, the given series becomes,
$\dfrac{x-1}{(x+1)}+\dfrac{1}{2}\cdot \dfrac{{{x}^{2}}-1}{{{(x+1)}^{2}}}+\dfrac{1}{3}\cdot \dfrac{{{x}^{3}}-1}{{{(x+1)}^{3}}}+.....\infty =-{{\log }_{e}}\left( 1-\dfrac{x}{x+1} \right)+{{\log }_{e}}\left( 1-\dfrac{1}{x+1} \right)$
On simplifying, we get
$\begin{align}
  & -{{\log }_{e}}\left( 1-\dfrac{x}{x+1} \right)+{{\log }_{e}}\left( 1-\dfrac{1}{x+1} \right) \\
 & \Rightarrow -{{\log }_{e}}\left( \dfrac{x+1-x}{x+1} \right)+{{\log }_{e}}\left( \dfrac{x+1-1}{x+1} \right) \\
 & \Rightarrow -{{\log }_{e}}\left( \dfrac{1}{x+1} \right)+{{\log }_{e}}\left( \dfrac{x}{x+1} \right) \\
 & \Rightarrow -{{\log }_{e}}(1)+{{\log }_{e}}(x+1)+{{\log }_{e}}(x)-{{\log }_{e}}(x+1) \\
 & \Rightarrow {{\log }_{e}}x \\
\end{align}$
Therefore, the sum is
$\dfrac{x-1}{(x+1)}+\dfrac{1}{2}\cdot \dfrac{{{x}^{2}}-1}{{{(x+1)}^{2}}}+\dfrac{1}{3}\cdot \dfrac{{{x}^{3}}-1}{{{(x+1)}^{3}}}+.....\infty ={{\log }_{e}}x$



Option ‘A’ is correct



Note: The given series is rewritten in order to get the appropriate progression in the series. Once the series represents a particular progression, we can find the sum of that series easily by applying the formula. Here the series is rewritten as the sum of two logarithmic series. So, by applying their formulae, we get the required sum. In such type of question, we need to observe the given series, so that we are able to find its progression. Then, the process of solving becomes easy for us.