Question

If we have a summation as ${{S}_{n}}=\sum\limits_{r=1}^{n}{\dfrac{r}{1.3.5......(2r+1)}}$ then?
(a) ${{S}_{n}}=\dfrac{1}{2}\left[ 1-\dfrac{r}{1.3.5......(2r+1)} \right]$
(b) ${{S}_{\infty }}=\dfrac{1}{2}$
(c) ${{S}_{n}}=\dfrac{1}{4}\left[ 1+\dfrac{r}{1.3.5......(2r+1)} \right]$
(d) ${{S}_{\infty }}=\dfrac{1}{4}$

Answer 1

Hint: We will start by substituting the value in the given expression. After that we will solve these questions by option. $\dfrac{\infty }{\infty }$ is an indeterminant form. So, use the L’ Hospital rule; in which we differentiate the numerator and the denominator separately i.e. if $\dfrac{f\left( x \right)}{g\left( x \right)}$ is our function, then after applying L’ Hospital rule it will become $\dfrac{\dfrac{d}{dx}f\left( x \right)}{\dfrac{d}{dx}g\left( x \right)}=\dfrac{f'\left( x \right)}{g'\left( x \right)}$.

Complete step-by-step solution:
According to the problem, we have given ${{S}_{n}}=\sum\limits_{r=1}^{n}{\dfrac{r}{1.3.5......(2r+1)}}$ and we need to find the values of ${{S}_{n}}$ and ${{S}_{\infty }}$.
Let us simplify the term $\dfrac{r}{1.3.5.7.......\left( 2r+1 \right)}$. Let us multiply numerator and denominator of this term with 2.
So, we get $\dfrac{r}{1.3.5.7.......\left( 2r+1 \right)}=\dfrac{1}{2}\times \dfrac{2r}{1.3.5.7.......\left( 2r+1 \right)}$.
$\Rightarrow \dfrac{r}{1.3.5.7.......\left( 2r+1 \right)}=\dfrac{1}{2}\times \left( \dfrac{2r+1-1}{1.3.5.7.......\left( 2r+1 \right)} \right)$.
$\Rightarrow \dfrac{r}{1.3.5.7.......\left( 2r-1 \right)}=\dfrac{1}{2}\times \left( \dfrac{2r+1}{1.3.5.7.......\left( 2r+1 \right)}-\dfrac{1}{1.3.5.7.......\left( 2r+1 \right)} \right)$.
$\Rightarrow \dfrac{r}{1.3.5.7.......\left( 2r-1 \right)}=\dfrac{1}{2}\times \left( \dfrac{1}{1.3.5.7.......\left( 2r-1 \right)}-\dfrac{1}{1.3.5.7.......\left( 2r+1 \right)} \right)$. Let us substitute this result in the required summation.
\[\Rightarrow {{S}_{n}}=\sum\limits_{r=1}^{n}{\left( \dfrac{1}{2}\times \left( \dfrac{1}{1.3.5.7.......\left( 2r-1 \right)}-\dfrac{1}{1.3.5.7.......\left( 2r+1 \right)} \right) \right)}\].
\[\Rightarrow {{S}_{n}}=\dfrac{1}{2}\sum\limits_{r=1}^{n}{\left( \dfrac{1}{1.3.5.7.......\left( 2r-1 \right)}-\dfrac{1}{1.3.5.7.......\left( 2r+1 \right)} \right)}\].
Let us substitute each term to get the summation.
\[\Rightarrow {{S}_{n}}=\dfrac{1}{2}\times \left( \dfrac{1}{1}-\dfrac{1}{1.3}+\dfrac{1}{1.3}-\dfrac{1}{1.3.5}+\dfrac{1}{1.3.5}-\dfrac{1}{1.3.5.7}+......+\dfrac{1}{1.3.5......\left( 2n-1 \right)}-\dfrac{1}{1.3.5......\left( 2n+1 \right)} \right)\].
We can see the second term is canceling the third term and the fourth term cancels the fifth term and this process continues till the first and last term remains.
So, we get \[{{S}_{n}}=\dfrac{1}{2}\times \left( 1-\dfrac{1}{1.3.5......\left( 2n+1 \right)} \right)\].
So, we have found the value of the sum of n-terms as \[{{S}_{n}}=\dfrac{1}{2}\times \left( 1-\dfrac{1}{1.3.5......\left( 2n+1 \right)} \right)\].
Now, let us substitute $\infty $ in place of n in the obtained sum.
So, we get \[{{S}_{\infty }}=\dfrac{1}{2}\times \left( 1-\dfrac{1}{1.3.5......\left( 2\left( \infty \right)+1 \right)} \right)\].
$\Rightarrow {{S}_{\infty }}=\dfrac{1}{2}\times \left( 1-\dfrac{1}{1.3.5......\infty } \right)$.
$\Rightarrow {{S}_{\infty }}=\dfrac{1}{2}\times \left( 1-\dfrac{1}{\infty } \right)$.
$\Rightarrow {{S}_{\infty }}=\dfrac{1}{2}\times \left( 1-0 \right)$.
$\Rightarrow {{S}_{\infty }}=\dfrac{1}{2}\times \left( 1 \right)$.
$\Rightarrow {{S}_{\infty }}=\dfrac{1}{2}$.
So, we have found the value of ${{S}_{\infty }}=\dfrac{1}{2}$.
Hence, the correct options will be (a) and (b).

Note: Since the absolute value of infinity is unknown we sometimes take the limit as infinity to find the value of ${{S}_{\infty }}$. We can also find the value of ${{S}_{1}}$ and cross verify the options by substituting 1 in place of n. We should see which terms are canceling in the series we just obtained after simplifying the term $\dfrac{r}{1.3.5.7.......\left( 2r+1 \right)}$. Here $\dfrac{1}{2}$ will be common for each term after expanding the summation so it can be taken outside of summation.