The value of ∑$$r=0$$∞ $$tan$$−$$1$$⁡$$11+r+r2$$ $$=$$πk then $$k=$$

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$The value of \sum_{r = 0}^{\infty} \tan^{- 1} (\frac{1}{1 + r + r^{2}}) = \frac{π}{k} then k=$

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Solution

$\tan^{- 1} (\frac{1}{1 + r + r^{2}}) = \tan^{- 1} (\frac{r + 1 - r}{1 + r (r + 1)})$
$= \tan^{- 1} (r + 1) - \tan^{- 1} (r)$
$\begin{array}{l} G i v e n q u e s t i o n \Rightarrow \sum_{r = 0}^{n} [\tan^{- 1} (r + 1) - \tan^{- 1} (r)] \\ = (\tan^{- 1} 1 - \tan^{- 1} 0) + (\tan^{- 1} 2 - \tan^{- 1} 1) + . . . . . . . . + (\tan^{- 1} (n + 1) - \tan^{- 1} (n)) \\ = \tan^{- 1} (n + 1) - \tan^{- 1} (0) \\ = \tan^{- 1} (n + 1) \\ \Rightarrow \lim_{n \to \infty} \sum_{r = 0}^{n} \tan^{- 1} (\frac{1}{1 + r + r^{2}}) = \tan^{- 1} (\infty) = \frac{π}{2} \end{array}$

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Properties of ITF

Remember concepts with our Masterclasses.

The value of ∑r=0∞ tan−1⁡11+r+r2 =πk then k=

Ready to Test Your Skills?

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$The value of \sum_{r = 0}^{\infty} \tan^{- 1} (\frac{1}{1 + r + r^{2}}) = \frac{π}{k} then k=$