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  4. Find the value of displaystyle∑n=1∞ tan -1 (1/1+n+n2)

Q. Find the value of $\displaystyle\sum_{n=1}^{\infty} \tan ^{-1} \frac{1}{1+n+n^{2}}$

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Solution:

$\sum \tan ^{-1} \frac{1}{1+n+n^{2}}=\sum \tan ^{-1}\left[\frac{n+1-n}{1+(n-1) n}\right]$
$=\sum\left[\tan ^{-1}(n+1)-\tan ^{-1} n\right]$
$=\left(\tan ^{-1} 2-\tan ^{-1} 1\right)+\left(\tan ^{-1} 3-\tan ^{-1} 2\right)+\ldots \ldots \ldots=-\tan ^{-1} 1+\tan ^{-1} \infty$
$=-\frac{\pi}{4}+\frac{\pi}{2}=\frac{\pi}{4}$

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