Q64 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics
$\lim _{n \rightarrow \infty} \sum_{r=1}^n \tan ^{-1}\left(\frac{2 r}{r^4+r^2+2}\right)=$
- Correct answerA$\frac{\pi}{4}$
- B$\frac{\pi}{2}$
- C$\frac{-\pi}{4}$
- D$\frac{-\pi}{2}$
Solution
Since,
$\begin{aligned}
& \quad r^4+r^2+1=\left(r^2-r+1\right)\left(r^2+r+1\right) \\
& \text { So, } \tan ^{-1}\left(\frac{2 r}{1+\left(r^4+r^2+1\right)}\right) \\
& =\tan ^{-1}\left(r^2+r+1\right)-\tan ^{-1}\left(r^2-r+1\right) \\
& \text { So, } \sum_{r=1}^n \tan ^{-1}\left(\frac{2 r}{r^4+r^2+2}\right) \\
& =\sum_{r=1}^n\left\{\tan ^{-1}\left(r^2+r+1\right)-\tan ^{-1}\left(r^2-r+1\right)\right\} \\
& =\tan ^{-1}\left(n^2+n+1\right)-\tan ^{-1}(1) \\
& =\tan ^{-1} \frac{n^2+n}{1+n^2+n+1}=\tan ^{-1} \frac{n^2+n}{n^2+n+2} \\
& \text { So, } \lim _{n \rightarrow \infty} \sum_{r=1}^n \tan ^{-1}\left(\frac{2 r}{r^4+r^2+2}\right) \\
& =\lim _{n \rightarrow \infty} \tan ^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)=\frac{\pi}{4} .
\end{aligned}$
Hence, the correct answer is option: 'A'.