Question

The of the series $1+\dfrac{1.3}{6}+\dfrac{1.3.5}{6.8}+............\infty $
$\begin{align}
  & \left( a \right)1 \\
 & \left( b \right)0 \\
 & \left( c \right)\infty \\
 & \left( d \right)4 \\
\end{align}$

Answer 1

Hint: To solve the question above, we will multiply and divide the series with that number so that when the terms are added, we would be able to cancel these terms. After doing this, we will find the general form of the above series and calculate the sum by the
 formula: \[{{S}_{n}}=\in {{T}_{n}}\]
Where ${{S}_{n}}$ is the sum of n form and ${{T}_{n}}$ is the general form of the series.

Complete step-by-step answer:
Before solving the question, we will multiply and divide the whole series with 8. Thus we will get the following series: $\Rightarrow S=\dfrac{8}{8}\left[ 1+\dfrac{1.3}{6}+\dfrac{1.3.5}{6.8}+..........\infty \right]$
                                    $\Rightarrow S=8\left[ \dfrac{1}{8}+\dfrac{1.3}{6.8}+\dfrac{1.3.5}{6.8.8}+.............\infty \right]$
                                   $\Rightarrow S=8\left[ \dfrac{1}{2.4}+\dfrac{1.3}{2.4.6}+\dfrac{1.3.5}{2.4.6.8}+...........\infty \right]$
Thus, the general terms of the above series can be represented by ${{T}_{n}}$ and it will be
given by: ${{T}_{n}}=\dfrac{1.3.5.7............\left( 2n-1 \right)}{2.4.6.8...........\left( 2n+2 \right)}$
Now, we can see that the number of terms which are in multiplication is numerator n and in denominator these terms are (n+1). To get (n+1) terms in numerator we will do
following realation: ${{T}_{n}}=\dfrac{1.3.5.7..........\left( 2n-1 \right)}{2.4.6.8..........\left( 2n+2 \right)}\left[ \left( 2n+2 \right)-\left( 2n+1 \right) \right]$
Now, we have to calculate the sum of all the terms in the series. Now, we know that if ${{S}_{n}}$ is the sum of all the terms till the${{n}^{th}}$ term and ${{T}_{n}}$ is the ${{n}^{th}}$form, then
 we will have the following relation: ${{S}_{n}}=\in _{n=1}^{n}{{T}_{n}}$
Now, we have to calculate the sum till the form reaches the $\infty $ i.e. the number of terms are$\infty $. Thus we will get: ${{S}_{n}}=\in _{n=1}^{\infty }{{T}_{n}}$
\[\Rightarrow {{S}_{\infty }}=8\left( {{T}_{1}}+{{T}_{2}}+{{T}_{3}}+{{T}_{4}}+..........{{T}_{\infty }} \right)\]
\[\Rightarrow {{S}_{\infty }}=8\left[ \begin{align}
  & \left( \dfrac{1\left[ 4-3 \right]}{2.4} \right)+\left( \dfrac{1.3\left[ 6-5 \right]}{2.4.6} \right)+\left( \dfrac{1.3.5\left[ 8-7 \right]}{2.4.6.8} \right)+ \\
 & {{\lim }_{n\to \infty }}\dfrac{1.3.5.7........\left( 2n-1 \right)}{2.4.6.8.........\left( 2n+2 \right)}\left[ \left( 2n+2 \right)-\left( 2n-1 \right) \right] \\
\end{align} \right]\]
\[\Rightarrow {{S}_{\infty }}=8\left[ \begin{align}
  & \left( \dfrac{1.4}{2.4}-\dfrac{1.3}{2.4} \right)+\left( \dfrac{1.3.6}{2.4.6}-\dfrac{1.3.5}{2.4.6} \right)+\left( \dfrac{1.3.5.8}{2.4.6.8}-\dfrac{1.3.5.7}{2.4.6.8} \right)+ \\
 & {{\lim }_{\to \infty }}\left[ \dfrac{1.3.5.7.........\left( 2n-1 \right)\left( 2n+2 \right)}{2.4.6..............\left( 2n+2 \right)}-\dfrac{1.3.5..............\left( 2n-1 \right)\left( 2n+1 \right)}{2.4.6.8.............\left( 2n \right)} \right] \\
\end{align} \right]\]
\[\Rightarrow {{S}_{\infty }}=8\left[ \begin{align}
  & \left( \dfrac{1.4}{2.4}-\dfrac{1.3}{2.4} \right)+\left( \dfrac{1.3}{2.4}-\dfrac{1.3.5}{2.4.6} \right)+\left( \dfrac{1.3.5}{2.4.6}-\dfrac{1.3.5.7}{2.4.6.8} \right)+ \\
 & {{\lim }_{\to \infty }}\left[ \dfrac{1.3.5.........\left( 2n-1 \right)}{2.4.6..............\left( 2n \right)}-\dfrac{1.3.5.9..........\left( 2n+1 \right)}{2.4.6.............\left( 2n+2 \right)} \right] \\
\end{align} \right]\]
Here, we can see that the forms are cancelling. The second and third are equal but opposite in sign. Similarly the fourth and fifth terms are equal but opposite is sign. Thus, all the terms will cancel but similarly except the first form and the last term. Thus, we will get:
$\begin{align}
  & \Rightarrow {{S}_{\infty }}=8\left[ \dfrac{1}{2}.\dfrac{4}{4}-{{\lim }_{n\to \infty }}\left[ \dfrac{1.3.5.7.9........\left( 2n+1 \right)}{2.4.6.8...........\left( 2n+2 \right)} \right] \right] \\
 & \Rightarrow {{S}_{\infty }}=8\left[ \dfrac{1}{2}-{{\lim }_{n\to \infty }}{{T}_{n+1}} \right] \\
 & \Rightarrow {{S}_{\infty }}=8\left[ \dfrac{1}{2}-0 \right] \\
 & \Rightarrow {{S}_{\infty }}=8\times \dfrac{1}{2} \\
 & \Rightarrow {{S}_{\infty }}=4 \\
\end{align}$
Hence, option (d) is correct.

Note: We have been able to calculate the above sum of series because the series given in the equation is a convergent series i.e. the ratio of the \[{{\left( n+1 \right)}^{th}}\] term to the ${{n}^{th}}$ term is less than 1. If the ratio had been greater than 1, then we would not be able to calculate the series.