Question

Find $\dfrac{2}{1}\cdot \dfrac{1}{3}+\dfrac{3}{2}\cdot \dfrac{1}{9}+\dfrac{4}{3}\cdot \dfrac{1}{27}+\dfrac{5}{4}\cdot \dfrac{1}{81}+...\infty =$
A.$\dfrac{1}{2}-{{\log }_{e}}\dfrac{2}{3}$
B. $-{{\log }_{e}}\dfrac{2}{3}$
C. $\dfrac{1}{2}+{{\log }_{e}}\left( \dfrac{2}{3} \right)$
D. None of these

Answer 1

Hint: In this question, we are to find the sum of the given series. Here the given series is a logarithmic series. So, by applying the appropriate formula, the required sum is to be calculated. In this question, we need to find its $nth$ term and then apply the logarithmic series formula, 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
${{\log }_{e}}(1-x)=-x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-...$
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 expansion is
$\dfrac{2}{1}\cdot \dfrac{1}{3}+\dfrac{3}{2}\cdot \dfrac{1}{9}+\dfrac{4}{3}\cdot \dfrac{1}{27}+\dfrac{5}{4}\cdot \dfrac{1}{81}+...\infty $
The $nth$ term of the above expansion is
${{t}_{n}}=\dfrac{n+1}{n}\cdot \dfrac{1}{{{3}^{n}}}$
So, we can write the given expansion as
$\dfrac{2}{1}\cdot \dfrac{1}{3}+\dfrac{3}{2}\cdot \dfrac{1}{9}+\dfrac{4}{3}\cdot \dfrac{1}{27}+\dfrac{5}{4}\cdot \dfrac{1}{81}+...\infty =\sum\limits_{n=1}^{\infty }{\left( \dfrac{n+1}{n}\cdot \dfrac{1}{{{3}^{n}}} \right)}$
On simplifying the obtained summation, we get
\[\begin{align}
  & \Rightarrow \sum\limits_{n=1}^{\infty }{\left( \dfrac{n-1}{n}\cdot \dfrac{1}{{{3}^{n}}} \right)} \\
 & \Rightarrow \sum\limits_{n=1}^{\infty }{\left( 1-\dfrac{1}{n} \right)}\dfrac{1}{{{3}^{n}}} \\
 & \Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{{{3}^{n}}}}-\sum\limits_{n=1}^{\infty }{\dfrac{1}{n\cdot {{3}^{n}}}}\text{ }...(1) \\
\end{align}\]
Since we know that, the first sum forms an infinite geometric series i.e.,
$\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{3}^{n}}}}=\dfrac{1}{3}+\dfrac{1}{{{3}^{2}}}+\dfrac{1}{{{3}^{3}}}+...\infty $
So, its sum to infinity is calculated by ${{S}_{\infty }}=\dfrac{a}{1-r}$
Thus, on substituting, we get
$\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{3}^{n}}}}=\dfrac{1}{3}+\dfrac{1}{{{3}^{2}}}+\dfrac{1}{{{3}^{3}}}+...\infty =\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}=\dfrac{1}{2}\text{ }...(2)$
And, the second sum forms an infinite logarithmic series i.e.,
\[\sum\limits_{n=1}^{\infty }{\dfrac{1}{n\cdot {{3}^{n}}}=\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( {}^{1}/{}_{3} \right)}^{n}}}{n}}}=\dfrac{1}{3}+\dfrac{{{\left( {}^{1}/{}_{3} \right)}^{2}}}{2}+\dfrac{{{\left( {}^{1}/{}_{3} \right)}^{3}}}{3}+....+\infty \]
Then, we can apply the formula:
$\begin{align}
  & {{\log }_{e}}(1-x)=-x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-... \\
 & \Rightarrow -{{\log }_{e}}(1-x)=x+\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+... \\
\end{align}$
So, we can write the logarithmic expansion as
$\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( {}^{1}/{}_{3} \right)}^{n}}}{n}}=-{{\log }_{e}}(1-\dfrac{1}{3})=-{{\log }_{e}}\left( \dfrac{2}{3} \right)\text{ }...(3)$
On substituting (2) and (3) in (4), we get
\[\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{3}^{n}}}}-\sum\limits_{n=1}^{\infty }{\dfrac{1}{n\cdot {{3}^{n}}}}=\dfrac{1}{2}-{{\log }_{e}}\left( \dfrac{2}{3} \right)\]
Therefore,
$\dfrac{2}{1}\cdot \dfrac{1}{3}+\dfrac{3}{2}\cdot \dfrac{1}{9}+\dfrac{4}{3}\cdot \dfrac{1}{27}+\dfrac{5}{4}\cdot \dfrac{1}{81}+...\infty =\dfrac{1}{2}-{{\log }_{e}}\left( \dfrac{2}{3} \right)$



Option ‘A’ is correct



Note: In this type of question we need to find the $nth$ term of the given series by observing the given terms. Then, we can able to find that it may be a logarithmic series. Otherwise, we do not understand. Once we know the $nth$ term, then we can find the sum by simply applying summation to the obtained $nth$ term for $n=1\to \infty $.