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$The sum \sum_{n = 1}^{\infty} \tan^{- 1} (\frac{1}{n^{2} + n + 1}) is equal to$

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π4

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-π2

π2

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Correct option is A

∑n=1∞ tan−1⁡1n2+n+1=∑n=1∞ tan−1⁡(n+1)−(n)1+(n+1)(n)=∑n=1∞ tan−1 (n+1)-tan-1n=tan-12-tan-11+tan-13-tan-12+.......tan-1∞=π2−tan−1⁡1=π4

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The sum ∑n=1∞ tan−1⁡1n2+n+1 is equal to

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$The sum \sum_{n = 1}^{\infty} \tan^{- 1} (\frac{1}{n^{2} + n + 1}) is equal to$