Question

The series \[\dfrac{1}{\left( n+1 \right)}+\dfrac{1}{2{{\left( n+1 \right)}^{2}}}+\dfrac{1}{3{{\left( n+1 \right)}^{3}}}+...\] has the same sum as the series
(a) $\dfrac{1}{n}-\dfrac{1}{2{{n}^{2}}}+\dfrac{1}{3{{n}^{3}}}-\dfrac{1}{4{{n}^{4}}}+...$
(b) $\dfrac{1}{n}+\dfrac{1}{2{{n}^{2}}}+\dfrac{1}{3{{n}^{3}}}-\dfrac{1}{4{{n}^{4}}}+...$
(c) $\dfrac{1}{n}+\dfrac{1}{{{2}^{2}}}.\dfrac{1}{{{n}^{2}}}+\dfrac{1}{{{2}^{3}}}.\dfrac{1}{{{n}^{3}}}+...$
(d) None of the above

Answer 1

Hint: We have to relate the given series with expansion of some function. So, we will consider two expansions as
$\begin{align}
  & \log \left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+\dfrac{{{x}^{5}}}{5}.............\left( i \right) \\
 & \log \left( 1-x \right)=-x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-\dfrac{{{x}^{5}}}{5}.............\left( ii \right) \\
\end{align}$

Complete step-by-step solution:
Now, we will proceed by substituting values of x as $x=\dfrac{1}{n+1}$ in (ii) and $x=\dfrac{1}{n}$ in (i) and then relate them.

Here, we have the series given:
\[{{S}_{n}}=\dfrac{1}{1\left( n+1 \right)}+\dfrac{1}{2{{\left( n+1 \right)}^{2}}}+\dfrac{1}{3{{\left( n+1 \right)}^{3}}}+...\]
Here, we need to observe the series and try to guess the relation with the expansions we know or any special series.
Here, we have two expansions:
$\begin{align}
  & \log \left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+\dfrac{{{x}^{5}}}{5}.............\left( i \right) \\
 & \log \left( 1-x \right)=-x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-\dfrac{{{x}^{5}}}{5}.............\left( ii \right) \\
\end{align}$
Now relating the given series with the second expansion:
$-\log \left( 1-x \right)=x+\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{4}}}{4}+\dfrac{{{x}^{5}}}{5}..........$
Now, if we put $x=\dfrac{1}{n+1}$ in both sides of the expansion:
$-\log \left( 1-\dfrac{1}{n+1} \right)=\dfrac{1}{n+1}+\dfrac{1}{2{{\left( n+1 \right)}^{2}}}+\dfrac{1}{3{{\left( n+1 \right)}^{3}}}+..........$
Hence, the given series is the expansion of \[-\log \left( 1-\dfrac{1}{n+1} \right)\]
\[\begin{align}
  & \Rightarrow -\log \left( \dfrac{n+1-1}{n+1} \right)=-\log \left( \dfrac{n}{n+1} \right) \\
 & \Rightarrow \log \left( \dfrac{n}{n+1} \right)\text{ }\left( -\log \dfrac{1}{X}=\log X \right) \\
 & \Rightarrow \log \left( 1+\dfrac{1}{n} \right)................\left( iii \right) \\
\end{align}\]
Now from expansion (i) of the series.
$\log \left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+\dfrac{{{x}^{5}}}{5}..........$
Put $x=\dfrac{1}{n}$, if we compare the above series with equation (iii)
Hence,
$\begin{align}
  & \log \left( 1+\dfrac{1}{n} \right)=\dfrac{1}{n}-\dfrac{1}{{{n}^{2}}\left( 2 \right)}+\dfrac{1}{{{n}^{3}}\left( 3 \right)}-\dfrac{1}{{{n}^{4}}\left( 4 \right)}+.......... \\
 & \log \left( 1+\dfrac{1}{n} \right)=\dfrac{1}{n}-\dfrac{1}{2{{n}^{2}}}+\dfrac{1}{3{{n}^{3}}}-\dfrac{1}{4{{n}^{4}}}+.......... \\
\end{align}$

Note: (i) One can simply waste lots of pages to check for any special series like AGP, or can be solved by creating $\left( {{T}_{r}}-{{T}_{r-1}} \right)$ difference in ${{r}^{th}}$ and ${{\left( r-1 \right)}^{th}}$ term. But it all will go in vain. In this question observation is the most important part and is the beauty of the question as well. Once observed then it will become very simple. One may not observe $\log \left( 1+x \right)\text{ or }\log \left( 1-x \right)$ expansions in one go because we have $'+'$ signs between the terms but in the expansions of $\log \left( 1+x \right)\text{ and }\log \left( 1-x \right)$; we have alternate negative signs and all terms have negative signs respectively.
Hence, one will look as given in solution.
(ii) At last once you get $-\log \left( \dfrac{n}{n+1} \right)$ ; one can write it as $-\left( \log n-\log \left( n+1 \right) \right)\text{ or }$$\log \left( n+1 \right)-\log n$ which will not give the required solution in form of expansion. Hence, change of $-\log \left( \dfrac{n}{n+1} \right)$ to $\log \left( \dfrac{n+1}{n} \right)$ and again to $\log \left( \dfrac{n}{n+1} \right)$ is the most important point of this question.