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Question: The sum to infinite terms of the series \({{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)+{{\sin }^...

The sum to infinite terms of the series sin⁡−1(12)+sin⁡−1(2−16)+sin⁡−1(3−223)+.....∞{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)+{{\sin }^{-1}}\left( \dfrac{\sqrt{2}-\sqrt{1}}{\sqrt{6}} \right)+{{\sin }^{-1}}\left( \dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}} \right)+.....\infty is equal to
(a) π8\dfrac{\pi }{8}
(b) π4\dfrac{\pi }{4}
(c) π2\dfrac{\pi }{2}
(d) π\pi

Explanation

Solution

e start solving the problem by making necessary arrangements in each term of the series. We then find the general form of all the terms present in the series. We then make the necessary calculations to get the general term in the form of sin⁡−1x−sin⁡−1y=sin⁡−1(x1−y2−y1−x2){{\sin }^{-1}}x-{{\sin }^{-1}}y={{\sin }^{-1}}\left( x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}} \right). We then use the simplified form of the general term in the sum of series and then make the necessary calculations to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the sum to infinite terms of the series sin⁡−1(12)+sin⁡−1(2−16)+sin⁡−1(3−223)+.....∞{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)+{{\sin }^{-1}}\left( \dfrac{\sqrt{2}-\sqrt{1}}{\sqrt{6}} \right)+{{\sin }^{-1}}\left( \dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}} \right)+.....\infty .
Let us assume S=sin⁡−1(12)+sin⁡−1(2−16)+sin⁡−1(3−223)+.....∞S={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)+{{\sin }^{-1}}\left( \dfrac{\sqrt{2}-\sqrt{1}}{\sqrt{6}} \right)+{{\sin }^{-1}}\left( \dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}} \right)+.....\infty .
⇒S=sin⁡−1(1−02)+sin⁡−1(2−16)+sin⁡−1(3−222×3)+.....∞\Rightarrow S={{\sin }^{-1}}\left( \dfrac{1-0}{\sqrt{2}} \right)+{{\sin }^{-1}}\left( \dfrac{\sqrt{2}-\sqrt{1}}{\sqrt{6}} \right)+{{\sin }^{-1}}\left( \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{{{2}^{2}}\times 3}} \right)+.....\infty .
⇒S=sin⁡−1(1−01×2)+sin⁡−1(2−12×3)+sin⁡−1(3−23×4)+.....∞\Rightarrow S={{\sin }^{-1}}\left( \dfrac{\sqrt{1}-\sqrt{0}}{\sqrt{1\times 2}} \right)+{{\sin }^{-1}}\left( \dfrac{\sqrt{2}-\sqrt{1}}{\sqrt{2\times 3}} \right)+{{\sin }^{-1}}\left( \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3\times 4}} \right)+.....\infty .
We can see that each term in the given sum resembles the form sin⁡−1(r−r−1r×(r+1)){{\sin }^{-1}}\left( \dfrac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r\times \left( r+1 \right)}} \right).
So, we get S=∑r=1∞sin⁡−1(r−r−1r×(r+1))S=\sum\limits_{r=1}^{\infty }{{{\sin }^{-1}}\left( \dfrac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r\times \left( r+1 \right)}} \right)}.
So, we get the general term as Tr=sin⁡−1(r−r−1r×(r+1)){{T}_{r}}={{\sin }^{-1}}\left( \dfrac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r\times \left( r+1 \right)}} \right).
Let us simplify the general term.
So, we get Tr=sin⁡−1(rr×(r+1)−r−1r×(r+1)){{T}_{r}}={{\sin }^{-1}}\left( \dfrac{\sqrt{r}}{\sqrt{r\times \left( r+1 \right)}}-\dfrac{\sqrt{r-1}}{\sqrt{r\times \left( r+1 \right)}} \right).
⇒Tr=sin⁡−1(1rrr+1−1r+1r−1r)\Rightarrow {{T}_{r}}={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{r}}\sqrt{\dfrac{r}{r+1}}-\dfrac{1}{\sqrt{r+1}}\sqrt{\dfrac{r-1}{r}} \right).
⇒Tr=sin⁡−1(1rr+1−1r+1−1r+1r−1r)\Rightarrow {{T}_{r}}={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{r}}\sqrt{\dfrac{r+1-1}{r+1}}-\dfrac{1}{\sqrt{r+1}}\sqrt{\dfrac{r-1}{r}} \right).
⇒Tr=sin⁡−1(1r1−1r+1−1r+11−1r)\Rightarrow {{T}_{r}}={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{r}}\sqrt{1-\dfrac{1}{r+1}}-\dfrac{1}{\sqrt{r+1}}\sqrt{1-\dfrac{1}{r}} \right).
⇒Tr=sin⁡−1(1r1−(1r+1)2−1r+11−(1r)2)\Rightarrow {{T}_{r}}={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{r}}\sqrt{1-{{\left( \dfrac{1}{\sqrt{r+1}} \right)}^{2}}}-\dfrac{1}{\sqrt{r+1}}\sqrt{1-{{\left( \dfrac{1}{\sqrt{r}} \right)}^{2}}} \right).
Let us assume that 1r=x\dfrac{1}{\sqrt{r}}=x and 1r+1=y\dfrac{1}{\sqrt{r+1}}=y ---(2).
⇒Tr=sin⁡−1(x1−y2−y1−x2)\Rightarrow {{T}_{r}}={{\sin }^{-1}}\left( x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}} \right).
We know that sin⁡−1x−sin⁡−1y=sin⁡−1(x1−y2−y1−x2){{\sin }^{-1}}x-{{\sin }^{-1}}y={{\sin }^{-1}}\left( x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}} \right).
⇒Tr=sin⁡−1(x)−sin⁡−1(y)\Rightarrow {{T}_{r}}={{\sin }^{-1}}\left( x \right)-{{\sin }^{-1}}\left( y \right) ---(3).
Let us substitute equation (2) in equation (3).
⇒Tr=sin⁡−1(1r)−sin⁡−1(1r+1)\Rightarrow {{T}_{r}}={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{r}} \right)-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{r+1}} \right) ---(4).
Let us substitute equation (4) in equation (3).
So, we get S=∑r=1∞(sin⁡−1(1r)−sin⁡−1(1r+1))S=\sum\limits_{r=1}^{\infty }{\left( {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{r}} \right)-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{r+1}} \right) \right)}.
Now, let us substitute each value of ‘r’.
⇒S=(sin⁡−1(11)−sin⁡−1(11+1))+(sin⁡−1(12)−sin⁡−1(12+1))+(sin⁡−1(13)−sin⁡−1(13+1))+......∞\Rightarrow S=\left( {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{1}} \right)-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{1+1}} \right) \right)+\left( {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2+1}} \right) \right)+\left( {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{3+1}} \right) \right)+......\infty .
⇒S=(sin⁡−1(11)−sin⁡−1(12))+(sin⁡−1(12)−sin⁡−1(13))+(sin⁡−1(13)−sin⁡−1(14))+......∞\Rightarrow S=\left( {{\sin }^{-1}}\left( \dfrac{1}{1} \right)-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right) \right)+\left( {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right) \right)+\left( {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{4}} \right) \right)+......\infty .
⇒S=sin⁡−1(1)\Rightarrow S={{\sin }^{-1}}\left( 1 \right) as the remaining terms will cancel each other and last term will be zero.
⇒S=π2\Rightarrow S=\dfrac{\pi }{2}.
So, we have found the sum to infinite terms of the series sin⁡−1(12)+sin⁡−1(2−16)+sin⁡−1(3−223)+.....∞{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)+{{\sin }^{-1}}\left( \dfrac{\sqrt{2}-\sqrt{1}}{\sqrt{6}} \right)+{{\sin }^{-1}}\left( \dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}} \right)+.....\infty as π2\dfrac{\pi }{2}.

So, the correct answer is “Option A”.

Note: Whenever we get this type of problem, we first try to find the general form of the terms present in the series as we can see that simplifying the general term provided half of the answer to us. We can also find the sum of ‘n’ terms first and then extend it to sum to infinity as shown below:
So, we have Tr=sin⁡−1(1r)−sin⁡−1(1r+1){{T}_{r}}={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{r}} \right)-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{r+1}} \right).
Now, we know that sum to ‘n’ terms is defined as Sn=∑r=1nTr=∑r=1n(sin⁡−1(1r)−sin⁡−1(1r+1)){{S}_{n}}=\sum\limits_{r=1}^{n}{{{T}_{r}}}=\sum\limits_{r=1}^{n}{\left( {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{r}} \right)-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{r+1}} \right) \right)}.
⇒Sn=(sin⁡−1(11)−sin⁡−1(12))+(sin⁡−1(12)−sin⁡−1(13))+....+(sin⁡−1(1n)−sin⁡−1(1n+1))\Rightarrow {{S}_{n}}=\left( {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{1}} \right)-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right) \right)+\left( {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right) \right)+....+\left( {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{n}} \right)-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{n+1}} \right) \right).
⇒Sn=sin⁡−1(1)−sin⁡−1(1n+1)\Rightarrow {{S}_{n}}={{\sin }^{-1}}\left( 1 \right)-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{n+1}} \right).
⇒Sn=π2−sin⁡−1(1n+1)\Rightarrow {{S}_{n}}=\dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{n+1}} \right).
Now, let us substitute ∞\infty in place of ‘n’ to get the sum to infinity of the series.
⇒S∞=π2−sin⁡−1(1∞+1)\Rightarrow {{S}_{\infty }}=\dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{\infty +1}} \right).
⇒S∞=π2−sin⁡−1(1∞)\Rightarrow {{S}_{\infty }}=\dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{1}{\infty } \right).
⇒S∞=π2−0\Rightarrow {{S}_{\infty }}=\dfrac{\pi }{2}-0.
⇒S∞=π2\Rightarrow {{S}_{\infty }}=\dfrac{\pi }{2}.