find Σ_(k=0) ^(+∞) arctan( (1/(k^2 +k+1))) .


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Question Number 28622 by abdo imad last updated on 28/Jan/18

$f i n d \sum_{k = 0}^{+ \infty} a r c t a n (\frac{1}{k^{2} + k + 1}) .$
Commented by abdo imad last updated on 30/Jan/18

$l e t p u t S_{n} = \sum_{k = 0}^{n} a r c t a n (\frac{1}{k^{2} + k + 1}) w e h a v e$ $\frac{1}{k^{2} + k + 1} = \frac{1}{1 + k (k + 1)} = \frac{k + 1 - k}{1 + k (k + 1)} l e t p u t k = t a n (u_{k})$ $= \frac{t a n (u_{k + 1}) - t a n (u_{k})}{1 + t a n (u_{k}) t a n (u_{k + 1})} = t a n (u_{k + 1} - u_{k}) \Rightarrow$ $a r t a n (\frac{1}{k^{2} + k + 1}) = u_{k + 1} - u_{k} a n d$ $S_{n} = \sum_{k = 0}^{n} (u_{k + 1} - u_{k}) = u_{1} - u_{0} + u_{2} - u_{1} + . . . . + u_{n + 1} - u_{n}$ $= u_{n + 1} - u_{0} = a r c t a n (n + 1) \Rightarrow {l i m}_{n \to + \infty} S_{n} = \frac{π}{4} .$
Commented by abdo imad last updated on 30/Jan/18

${l i m}_{n \to + \infty} S_{n} = \frac{π}{2} .$
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