Series Inverse Sine Value Calculation

The problem asks for the sum of an infinite series involving inverse sine functions. The series is given by:

$$ \sin^{-1} \dfrac{1}{\sqrt{2}}+ \sin^{-1} \dfrac{\sqrt{2}-\sqrt{1}}{\sqrt{6}} + \sin^{-1} \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+...\;\text{upto infinity} $$

We need to find the value of this sum.

Identifying Inverse Sine Series Pattern

Let's examine the terms of the series:

• Term 1: \( \sin^{-1} \dfrac{1}{\sqrt{2}} \)
• Term 2: \( \sin^{-1} \dfrac{\sqrt{2}-1}{\sqrt{6}} \)
• Term 3: \( \sin^{-1} \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{12}} \)

We can try to express these terms using the inverse trigonometric identity: $$ \sin^{-1} x - \sin^{-1} y = \sin^{-1} \left( x\sqrt{1-y^2} - y\sqrt{1-x^2} \right) $$

Let's consider the sequence \( a_n = \sin^{-1} \dfrac{1}{\sqrt{n}} \).

Let's check the difference \( a_n - a_{n+1} \):

For \( n=1 \): $$ a_1 - a_2 = \sin^{-1} \dfrac{1}{\sqrt{1}} - \sin^{-1} \dfrac{1}{\sqrt{2}} = \sin^{-1} 1 - \sin^{-1} \dfrac{1}{\sqrt{2}} $$ Using the identity: $$ \sin^{-1} 1 \cdot \sqrt{1 - \left(\dfrac{1}{\sqrt{2}}\right)^2} - \sin^{-1} \dfrac{1}{\sqrt{2}} \cdot \sqrt{1 - 1^2} $$ $$ = \sin^{-1} \left( 1 \cdot \sqrt{1 - \dfrac{1}{2}} - \dfrac{1}{\sqrt{2}} \cdot \sqrt{0} \right) = \sin^{-1} \left( \sqrt{\dfrac{1}{2}} \right) = \sin^{-1} \dfrac{1}{\sqrt{2}} $$ This matches the first term of the given series.

For \( n=2 \): $$ a_2 - a_3 = \sin^{-1} \dfrac{1}{\sqrt{2}} - \sin^{-1} \dfrac{1}{\sqrt{3}} $$ Using the identity: $$ \sin^{-1} \left( \dfrac{1}{\sqrt{2}} \sqrt{1 - \left(\dfrac{1}{\sqrt{3}}\right)^2} - \dfrac{1}{\sqrt{3}} \sqrt{1 - \left(\dfrac{1}{\sqrt{2}}\right)^2} \right) $$ $$ = \sin^{-1} \left( \dfrac{1}{\sqrt{2}} \sqrt{1 - \dfrac{1}{3}} - \dfrac{1}{\sqrt{3}} \sqrt{1 - \dfrac{1}{2}} \right) $$ $$ = \sin^{-1} \left( \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{3}} - \dfrac{1}{\sqrt{3}} \cdot \dfrac{1}{\sqrt{2}} \right) $$ $$ = \sin^{-1} \left( \dfrac{1}{\sqrt{3}} - \dfrac{1}{\sqrt{6}} \right) = \sin^{-1} \left( \dfrac{\sqrt{2}}{\sqrt{6}} - \dfrac{1}{\sqrt{6}} \right) = \sin^{-1} \left( \dfrac{\sqrt{2}-1}{\sqrt{6}} \right) $$ This matches the second term of the given series.

For \( n=3 \): $$ a_3 - a_4 = \sin^{-1} \dfrac{1}{\sqrt{3}} - \sin^{-1} \dfrac{1}{\sqrt{4}} $$ Using the identity: $$ \sin^{-1} \left( \dfrac{1}{\sqrt{3}} \sqrt{1 - \left(\dfrac{1}{\sqrt{4}}\right)^2} - \dfrac{1}{\sqrt{4}} \sqrt{1 - \left(\dfrac{1}{\sqrt{3}}\right)^2} \right) $$ $$ = \sin^{-1} \left( \dfrac{1}{\sqrt{3}} \sqrt{1 - \dfrac{1}{4}} - \dfrac{1}{2} \sqrt{1 - \dfrac{1}{3}} \right) $$ $$ = \sin^{-1} \left( \dfrac{1}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{2} - \dfrac{1}{2} \cdot \dfrac{\sqrt{2}}{\sqrt{3}} \right) $$ $$ = \sin^{-1} \left( \dfrac{1}{2} - \dfrac{\sqrt{2}}{2\sqrt{3}} \right) = \sin^{-1} \left( \dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}} \right) = \sin^{-1} \left( \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{12}} \right) $$ This matches the third term of the given series.

Thus, the series can be written as: $$ \sum_{n=1}^{\infty} \left( \sin^{-1} \dfrac{1}{\sqrt{n}} - \sin^{-1} \dfrac{1}{\sqrt{n+1}} \right) $$

Sum Calculation for Telescoping Series

This is a telescoping series. Let \( S_N \) be the sum of the first \( N \) terms:

$$ S_N = \sum_{n=1}^{N} \left( \sin^{-1} \dfrac{1}{\sqrt{n}} - \sin^{-1} \dfrac{1}{\sqrt{n+1}} \right) $$

Expanding the sum: $$ S_N = \left( \sin^{-1} \dfrac{1}{\sqrt{1}} - \sin^{-1} \dfrac{1}{\sqrt{2}} \right) + \left( \sin^{-1} \dfrac{1}{\sqrt{2}} - \sin^{-1} \dfrac{1}{\sqrt{3}} \right) + \dots + \left( \sin^{-1} \dfrac{1}{\sqrt{N}} - \sin^{-1} \dfrac{1}{\sqrt{N+1}} \right) $$

Most terms cancel out, leaving: $$ S_N = \sin^{-1} \dfrac{1}{\sqrt{1}} - \sin^{-1} \dfrac{1}{\sqrt{N+1}} $$ $$ S_N = \sin^{-1} 1 - \sin^{-1} \dfrac{1}{\sqrt{N+1}} $$ $$ S_N = \dfrac{\pi}{2} - \sin^{-1} \dfrac{1}{\sqrt{N+1}} $$

Infinite Sum of Inverse Sine Series

To find the sum of the infinite series, we take the limit of \( S_N \) as \( N \) approaches infinity:

$$ S = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left( \dfrac{\pi}{2} - \sin^{-1} \dfrac{1}{\sqrt{N+1}} \right) $$

As \( N \to \infty \), \( \dfrac{1}{\sqrt{N+1}} \to 0 \). Therefore, $$ \lim_{N \to \infty} \sin^{-1} \dfrac{1}{\sqrt{N+1}} = \sin^{-1} 0 = 0 $$

So, the sum of the series is: $$ S = \dfrac{\pi}{2} - 0 = \dfrac{\pi}{2} $$