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$\large \sin^{-1}\dfrac{1}{\sqrt{2}} + \sin^{-1}\dfrac{\sqrt{2} - 1}{\sqrt{6}} + \sin^{-1}\dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{12}} + \cdots= \ ?$ 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

Note by Akhil Bansal
9 months, 3 weeks ago

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$\sum_{n=1}^{\infty}sin^{-1}(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}})$ 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 $\sum_{n=1}^{\infty}sin^{-1}(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}})= \sum_{n=1}^{\infty} tan^{-1}(\frac{\sqrt{n}-\sqrt{n-1}}{1+\sqrt{n(n-1)}})$ $= \sum_{n=1}^{\infty} tan^{-1}(\sqrt{n})-tan^{-1}(\sqrt{n-1})$ $= tan^{-1}(\infty)-tan^{-1}(0)$ $=\frac{\pi}{2}$. · 9 months, 3 weeks ago

nice one. dont give $$\infty$$ in an expression like that. arithmetic for $$\infty$$ is not defined. write as $\lim_{n\rightarrow \infty} (\tan^{-1}(n)-\tan^{-1}(0))$ · 9 months, 2 weeks ago

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 · 9 months, 3 weeks ago

Comment deleted 9 months ago

What is the general term to given expression? · 9 months, 3 weeks ago

@Akhil Bansal The general term is $$\sin^{-1} \left( \dfrac{\sqrt n -\sqrt{n-1}}{\sqrt{n(n+1)}} \right)$$. · 9 months, 3 weeks ago