\[\large \sin^{-1}\dfrac{1}{\sqrt{2}} + \sin^{-1}\dfrac{\sqrt{2} - 1}{\sqrt{6}} + \sin^{-1}\dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{12}} + \cdots= \ ? \] I am not able to generalise the series..

This is a MCQ type question and options are :

- 0
- 1
- \(\pi / 2\)
- 2
- None of these

## Comments

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TopNewest\[\sum_{n=1}^{\infty}sin^{-1}(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}})\] The other side in a right angled Triangle with other sides \(\sqrt{n(n+1)},\sqrt{n}-\sqrt{n-1}\) is \(\sqrt{n^{2}-n}+1\) Therefore \[\sum_{n=1}^{\infty}sin^{-1}(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}})= \sum_{n=1}^{\infty} tan^{-1}(\frac{\sqrt{n}-\sqrt{n-1}}{1+\sqrt{n(n-1)}})\] \[= \sum_{n=1}^{\infty} tan^{-1}(\sqrt{n})-tan^{-1}(\sqrt{n-1})\] \[= tan^{-1}(\infty)-tan^{-1}(0)\] \[=\frac{\pi}{2}\]. – Shivam Jadhav · 1 year, 3 months ago

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– Aareyan Manzoor · 1 year, 3 months ago

nice one. dont give \(\infty\) in an expression like that. arithmetic for \(\infty\) is not defined. write as \[\lim_{n\rightarrow \infty} (\tan^{-1}(n)-\tan^{-1}(0))\]Log in to reply

Notist that sin(arcsin(1/√2)+arcsin((√2-1)/√6)= 1/√2

(√2-1)/√6+1/√2(√2+1)/√6= 2√2/√12=√(2/3) So sum first two terms is arcsin(√(2/3)) Now sin(arcsin(√(2/3)+arcsin((√3-√2)/√12)=√(2/3)(√6+1)/√12+ +√(1/3)(√3-√2)/√12= 1/√36*(√12+√2+√3-√2)=3√3/6= √(3/4) so using mathematical induction you can prove that sum of first n terms is arccos(√(n/(n+1))= arccos(√(1-1/(n+1)) So when you put lim n->infinify You'll get arcsin(1)=π/2 – Nikola Djuric · 1 year, 3 months agoLog in to reply

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– Akhil Bansal · 1 year, 3 months ago

What is the general term to given expression?Log in to reply

@Akhil Bansal The general term is \(\sin^{-1} \left( \dfrac{\sqrt n -\sqrt{n-1}}{\sqrt{n(n+1)}} \right) \). – Sandeep Bhardwaj · 1 year, 3 months ago

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– Akhil Bansal · 1 year, 3 months ago

Thanks Sir, now I will do the rest.Log in to reply