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JEE Advanced 2022 Paper 2 · Mathematics · Trigonometry and Inverse Trigonometric Functions · hard

Let Sn=Σk=1^ntan ^-1 ((2)/(k^2). Then lim n→∞n((3π)/(4)-Sn) equals

Single correct option

Let Sn=∑k=1ntan⁡−1 ⁣(2k2)S_n=\displaystyle\sum_{k=1}^{n}\tan^{-1}\!\left(\frac{2}{k^{2}}\right). Then lim⁡n→∞n(3π4−Sn)\displaystyle\lim_{n\to\infty}n\left(\frac{3\pi}{4}-S_n\right) equals
  1. A. 1
  2. B. 2
  3. C. 4
  4. D. π\pi
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Answer: (B) 2

Since tan⁡(tan⁡−1(k+1)−tan⁡−1(k−1))=21+(k2−1)=2k2\tan\bigl(\tan^{-1}(k+1)-\tan^{-1}(k-1)\bigr)=\dfrac{2}{1+(k^2-1)}=\dfrac2{k^2} and both angles are in (−π2,π2)(-\frac\pi2,\frac\pi2) with a difference in [0,π2)[0,\frac\pi2), we get tan⁡−12k2=tan⁡−1(k+1)−tan⁡−1(k−1)\tan^{-1}\frac2{k^2}=\tan^{-1}(k+1)-\tan^{-1}(k-1). The sum telescopes: Sn=tan⁡−1n+tan⁡−1(n+1)−tan⁡−11−tan⁡−10=tan⁡−1n+tan⁡−1(n+1)−π4S_n=\tan^{-1}n+\tan^{-1}(n+1)-\tan^{-1}1-\tan^{-1}0=\tan^{-1}n+\tan^{-1}(n+1)-\frac\pi4. Hence Sn→3π4S_n\to\frac{3\pi}4. Using tan⁡−1m=π2−tan⁡−11m\tan^{-1}m=\frac\pi2-\tan^{-1}\frac1m, we get 3π4−Sn=tan⁡−11n+tan⁡−11n+1\frac{3\pi}4-S_n=\tan^{-1}\frac1n+\tan^{-1}\frac1{n+1}, so n(3π4−Sn)→1+1=2n\left(\frac{3\pi}4-S_n\right)\to1+1=2. Numerically, n=105n=10^5 gives 1.999991.99999.
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