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$\lim _{n \rightarrow \infty} \sum_{r=1}^n \tan ^{-1}\left(\frac{2 r}{r^4+r^2+2}\right)=$
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Verified Answer
The correct answer is:
$\frac{\pi}{4}$
Since,
$\begin{aligned}
& \quad r^4+r^2+1=\left(r^2-r+1\right)\left(r^2+r+1\right) \\
& \text { So, } \tan ^{-1}\left(\frac{2 r}{1+\left(r^4+r^2+1\right)}\right) \\
& =\tan ^{-1}\left(r^2+r+1\right)-\tan ^{-1}\left(r^2-r+1\right) \\
& \text { So, } \sum_{r=1}^n \tan ^{-1}\left(\frac{2 r}{r^4+r^2+2}\right) \\
& =\sum_{r=1}^n\left\{\tan ^{-1}\left(r^2+r+1\right)-\tan ^{-1}\left(r^2-r+1\right)\right\} \\
& =\tan ^{-1}\left(n^2+n+1\right)-\tan ^{-1}(1) \\
& =\tan ^{-1} \frac{n^2+n}{1+n^2+n+1}=\tan ^{-1} \frac{n^2+n}{n^2+n+2} \\
& \text { So, } \lim _{n \rightarrow \infty} \sum_{r=1}^n \tan ^{-1}\left(\frac{2 r}{r^4+r^2+2}\right) \\
& =\lim _{n \rightarrow \infty} \tan ^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)=\frac{\pi}{4} .
\end{aligned}$
$\begin{aligned}
& \quad r^4+r^2+1=\left(r^2-r+1\right)\left(r^2+r+1\right) \\
& \text { So, } \tan ^{-1}\left(\frac{2 r}{1+\left(r^4+r^2+1\right)}\right) \\
& =\tan ^{-1}\left(r^2+r+1\right)-\tan ^{-1}\left(r^2-r+1\right) \\
& \text { So, } \sum_{r=1}^n \tan ^{-1}\left(\frac{2 r}{r^4+r^2+2}\right) \\
& =\sum_{r=1}^n\left\{\tan ^{-1}\left(r^2+r+1\right)-\tan ^{-1}\left(r^2-r+1\right)\right\} \\
& =\tan ^{-1}\left(n^2+n+1\right)-\tan ^{-1}(1) \\
& =\tan ^{-1} \frac{n^2+n}{1+n^2+n+1}=\tan ^{-1} \frac{n^2+n}{n^2+n+2} \\
& \text { So, } \lim _{n \rightarrow \infty} \sum_{r=1}^n \tan ^{-1}\left(\frac{2 r}{r^4+r^2+2}\right) \\
& =\lim _{n \rightarrow \infty} \tan ^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)=\frac{\pi}{4} .
\end{aligned}$
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