Please help!

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

#Geometry

Easy Math Editor

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Comments

$\sum_{n=1}^{\infty}sin^{-1}(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}})$ 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 - 5 years, 6 months ago

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

Aareyan Manzoor - 5 years, 6 months 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

Nikola Djuric - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$