Exams › JEE Main › Maths
Evaluate the value of the infinite sum \(\sin^{-1}\!\left(\frac{1}{\sqrt{2}}\right)+\sin^{-1}\!\left(\frac{\sqrt{2}-1}{\sqrt{6}}\right)+\sin^{-1}\!\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}\right)+\cdots+\sin^{-1}\!\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}}\right)+\cdots\).
- \(\pi/8\)
- \(\pi/4\)
- \(\pi/2\)
- \(\pi\)
Correct answer: \(\pi/2\)
Solution
The infinite sum converges to \\frac{C0}{2} because each term in the series represents the inverse sine of a value that approaches the limit of 1 as n increases, leading to the cumulative angle approaching \\frac{C0}{2}.
Related JEE Main Maths questions
⚔️ Practice JEE Main Maths free + battle 1v1 →