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  4. displaystyle lim n arrow ∞ tan displaystyle∑r=1n tan -1((1/1+r+r2)) is equal to .

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Q. $\displaystyle\lim _{n \rightarrow \infty} \tan \left\{\displaystyle\sum_{r=1}^{n} \tan ^{-1}\left(\frac{1}{1+r+r^{2}}\right)\right\}$ is equal to _______.

JEE MainJEE Main 2021Inverse Trigonometric Functions

Solution:

$\displaystyle\lim _{n \rightarrow a} \tan \left(\displaystyle\sum_{r=1}^{n} \tan ^{-1}\left(\frac{1}{1+r(r+1)}\right)\right)$
$=\displaystyle\lim _{n \rightarrow \alpha} \tan \left(\displaystyle\sum_{r=1}^{n} \tan ^{-1}\left(\frac{r+1-r}{1+r(r+1)}\right)\right)$
$=\tan \left(\displaystyle\lim _{n \rightarrow a} \displaystyle\sum_{r=1}^{n}\left[\tan ^{-1}(r+1)-\tan ^{-1}(r)\right]\right)$
$=\tan \left(\displaystyle\lim _{n \rightarrow \infty}\left(\tan ^{-1}(n+1)-\frac{\pi}{4}\right)\right)$
$=\tan \left(\frac{\pi}{4}\right)=1$