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Question: If we have a summation as \({{S}_{n}}=\sum\limits_{r=1}^{n}{\dfrac{r}{1.3.5......(2r+1)}}\) then? ...

If we have a summation as Sn=∑r=1nr1.3.5......(2r+1){{S}_{n}}=\sum\limits_{r=1}^{n}{\dfrac{r}{1.3.5......(2r+1)}} then?
(a) Sn=12[1−r1.3.5......(2r+1)]{{S}_{n}}=\dfrac{1}{2}\left[ 1-\dfrac{r}{1.3.5......(2r+1)} \right]
(b) S∞=12{{S}_{\infty }}=\dfrac{1}{2}
(c) Sn=14[1+r1.3.5......(2r+1)]{{S}_{n}}=\dfrac{1}{4}\left[ 1+\dfrac{r}{1.3.5......(2r+1)} \right]
(d) S∞=14{{S}_{\infty }}=\dfrac{1}{4}

Explanation

Solution

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 f(x)g(x)\dfrac{f\left( x \right)}{g\left( x \right)} is our function, then after applying L’ Hospital rule it will become ddxf(x)ddxg(x)=f′(x)g′(x)\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 Sn=∑r=1nr1.3.5......(2r+1){{S}_{n}}=\sum\limits_{r=1}^{n}{\dfrac{r}{1.3.5......(2r+1)}} and we need to find the values of Sn{{S}_{n}} and S∞{{S}_{\infty }}.
Let us simplify the term r1.3.5.7.......(2r+1)\dfrac{r}{1.3.5.7.......\left( 2r+1 \right)}. Let us multiply numerator and denominator of this term with 2.
So, we get r1.3.5.7.......(2r+1)=12×2r1.3.5.7.......(2r+1)\dfrac{r}{1.3.5.7.......\left( 2r+1 \right)}=\dfrac{1}{2}\times \dfrac{2r}{1.3.5.7.......\left( 2r+1 \right)}.
⇒r1.3.5.7.......(2r+1)=12×(2r+1−11.3.5.7.......(2r+1))\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).
⇒r1.3.5.7.......(2r−1)=12×(2r+11.3.5.7.......(2r+1)−11.3.5.7.......(2r+1))\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).
⇒r1.3.5.7.......(2r−1)=12×(11.3.5.7.......(2r−1)−11.3.5.7.......(2r+1))\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.
⇒Sn=∑r=1n(12×(11.3.5.7.......(2r−1)−11.3.5.7.......(2r+1)))\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)}.
⇒Sn=12∑r=1n(11.3.5.7.......(2r−1)−11.3.5.7.......(2r+1))\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.
⇒Sn=12×(11−11.3+11.3−11.3.5+11.3.5−11.3.5.7+......+11.3.5......(2n−1)−11.3.5......(2n+1))\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 Sn=12×(1−11.3.5......(2n+1)){{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 Sn=12×(1−11.3.5......(2n+1)){{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∞=12×(1−11.3.5......(2(∞)+1)){{S}_{\infty }}=\dfrac{1}{2}\times \left( 1-\dfrac{1}{1.3.5......\left( 2\left( \infty \right)+1 \right)} \right).
⇒S∞=12×(1−11.3.5......∞)\Rightarrow {{S}_{\infty }}=\dfrac{1}{2}\times \left( 1-\dfrac{1}{1.3.5......\infty } \right).
⇒S∞=12×(1−1∞)\Rightarrow {{S}_{\infty }}=\dfrac{1}{2}\times \left( 1-\dfrac{1}{\infty } \right).
⇒S∞=12×(1−0)\Rightarrow {{S}_{\infty }}=\dfrac{1}{2}\times \left( 1-0 \right).
⇒S∞=12×(1)\Rightarrow {{S}_{\infty }}=\dfrac{1}{2}\times \left( 1 \right).
⇒S∞=12\Rightarrow {{S}_{\infty }}=\dfrac{1}{2}.
So, we have found the value of S∞=12{{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∞{{S}_{\infty }}. We can also find the value of S1{{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 r1.3.5.7.......(2r+1)\dfrac{r}{1.3.5.7.......\left( 2r+1 \right)}. Here 12\dfrac{1}{2} will be common for each term after expanding the summation so it can be taken outside of summation.