Sine of Summations!

Algebra Level 4

∑ k = 1 ∞ sin ⁡ − 1 ( k − k − 1 k ( k + 1 ) ) = θ \large \sum _{ k=1 }^{ \infty }{ \sin ^{ -1 }{ \left( \frac { \sqrt { k } - \sqrt { k - 1 } }{ \sqrt { k( k + 1) } } \right) } } = \theta

Find sin ⁡ θ \sin \theta .


The answer is 1.

Priyanshu Mishra by Priyanshu Mishra , 20, India

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1 solution

Mark Hennings
Feb 14, 2018

Since k ( k + 1 ) − ( k − k − 1 ) 2 = k 2 − k + 1 + 2 k ( k − 1 ) = ( k ( k − 1 ) + 1 ) 2 k(k+1) - (\sqrt{k} - \sqrt{k-1})^2 \; = \; k^2 - k + 1 + 2\sqrt{k(k-1)} \; = \; \big(\sqrt{k(k-1)} + 1\big)^2 we deduce that sin ⁡ − 1 ( k − k − 1 k ( k + 1 ) ) = tan ⁡ − 1 ( k − k − 1 1 + k ( k − 1 ) ) = tan ⁡ − 1 k − tan ⁡ − 1 k − 1 \sin^{-1}\left(\frac{\sqrt{k} - \sqrt{k-1}}{\sqrt{k(k+1)}}\right) \; = \; \tan^{-1}\left(\frac{\sqrt{k} - \sqrt{k-1}}{1 + \sqrt{k(k-1)}}\right) \; = \; \tan^{-1}\sqrt{k} - \tan^{-1}\sqrt{k-1} and hence the series telescopes, with θ = ∑ k = 1 ∞ ( tan ⁡ − 1 k − tan ⁡ − 1 k − 1 ) = 1 2 π \theta \; = \; \sum_{k=1}^\infty \big(\tan^{-1}\sqrt{k} - \tan^{-1}\sqrt{k-1}\big) \; = \; \tfrac12\pi which makes the answer 1 \boxed{1} .

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