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JEE Advanced 2024 Paper 1 · Mathematics · Trigonometry and Inverse Trigonometric Functions · hard

Let S=Σk=1^∞tan ^-1 ((2)/(k^2), where tan ^-1 takes values in (-(π)/(2),(π)/(2)). The value of (4S)/(π) is ______.

Numerical answer

Let S=∑k=1∞tan⁡−1 ⁣(2k2)S=\displaystyle\sum_{k=1}^{\infty}\tan^{-1}\!\left(\dfrac{2}{k^{2}}\right), where tan⁡−1\tan^{-1} takes values in (−π2,π2)\left(-\tfrac\pi2,\tfrac\pi2\right). The value of 4Sπ\dfrac{4S}{\pi} is ______.
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Answer: 3

Use tan⁡−1a−tan⁡−1b=tan⁡−1a−b1+ab\tan^{-1}a-\tan^{-1}b=\tan^{-1}\dfrac{a-b}{1+ab} (valid here since ab>−1ab>-1). Take a=k+1a=k+1, b=k−1b=k-1: a−b1+ab=21+(k2−1)=2k2\dfrac{a-b}{1+ab}=\dfrac{2}{1+(k^2-1)}=\dfrac{2}{k^2}. Hence tan⁡−12k2=tan⁡−1(k+1)−tan⁡−1(k−1)\tan^{-1}\dfrac{2}{k^2}=\tan^{-1}(k+1)-\tan^{-1}(k-1) for every k≥1k\ge1. The partial sum telescopes: ∑k=1N=tan⁡−1(N+1)+tan⁡−1N−tan⁡−11−tan⁡−10\sum_{k=1}^{N}=\tan^{-1}(N+1)+\tan^{-1}N-\tan^{-1}1-\tan^{-1}0. As N→∞N\to\infty this tends to π2+π2−π4=3π4\dfrac\pi2+\dfrac\pi2-\dfrac\pi4=\dfrac{3\pi}{4}. So S=3π4S=\dfrac{3\pi}4 and 4Sπ=3\dfrac{4S}{\pi}=3. A numerical summation of 200000 terms gives 4S/π=2.999994S/\pi=2.99999, consistent with 3.
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