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Question Number 165349 by mkam last updated on 31/Jan/22

Answered by puissant last updated on 31/Jan/22

(1/(2n^2 )) = (((2n+1)−(2n−1))/(1+(2n+1)(2n−1))) ⇒Σ_(n=1) ^∞ arctan((1/(2n^2 )))=Σ_(n=1) ^∞ {arctan(2n+1)−arctan(2n−1)} ⇒ Σ_(n=1) ^∞ arctan((1/(2n^2 )))=lim_(n→∞) Σ_(k=1) ^n {arctan(2k+1)−arctan(2k−1)} = lim_(n→∞) (−arctan(1)+arctan(2n+1)) ( somme telescopique). = −(π/4)+(π/2) = (π/4). ...............Le puissant............

$$\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }\:=\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)−\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{1}+\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)} \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left\{{arctan}\left(\mathrm{2}{n}+\mathrm{1}\right)−{arctan}\left(\mathrm{2}{n}−\mathrm{1}\right)\right\} \\ $$$$\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left\{{arctan}\left(\mathrm{2}{k}+\mathrm{1}\right)−{arctan}\left(\mathrm{2}{k}−\mathrm{1}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(−{arctan}\left(\mathrm{1}\right)+{arctan}\left(\mathrm{2}{n}+\mathrm{1}\right)\right) \\ $$$$\left(\:{somme}\:{telescopique}\right). \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{2}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\pi}{\mathrm{4}}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...............\mathscr{L}{e}\:{puissant}............ \\ $$

Commented by puissant last updated on 31/Jan/22

=lim_(k→∞) Σ_(t=1) ^k arctan(2t+1)−arcan(2t−1) = lim_(t→∞) [arcan(3)−arctan(1)+arctan(5)−arctan(3)+ arctan(7)−arctan(5)....+arctan(2k+1)−arctan(2k−1)] = lim_(k→∞) {−arctan(1)+arctan(2k+1)} = −(π/4)+(π/2) = (π/4).

$$=\underset{{k}\rightarrow\infty} {\mathrm{lim}}\underset{{t}=\mathrm{1}} {\overset{{k}} {\sum}}{arctan}\left(\mathrm{2}{t}+\mathrm{1}\right)−{arcan}\left(\mathrm{2}{t}−\mathrm{1}\right) \\ $$$$=\:\underset{{t}\rightarrow\infty} {\mathrm{lim}}\left[{arcan}\left(\mathrm{3}\right)−{arctan}\left(\mathrm{1}\right)+{arctan}\left(\mathrm{5}\right)−{arctan}\left(\mathrm{3}\right)+\right. \\ $$$$\left.{arctan}\left(\mathrm{7}\right)−{arctan}\left(\mathrm{5}\right)....+{arctan}\left(\mathrm{2}{k}+\mathrm{1}\right)−{arctan}\left(\mathrm{2}{k}−\mathrm{1}\right)\right] \\ $$$$=\:\underset{{k}\rightarrow\infty} {\mathrm{lim}}\left\{−{arctan}\left(\mathrm{1}\right)+{arctan}\left(\mathrm{2}{k}+\mathrm{1}\right)\right\} \\ $$$$=\:−\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{2}}\:=\:\frac{\pi}{\mathrm{4}}. \\ $$

Commented by mkam last updated on 31/Jan/22

sir how Σ_(k=1) ^∞ {arctan(2k+1)−arctan(2k−1)}= −arctan(1)+arctan(2n+1)

$${sir}\:{how}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left\{{arctan}\left(\mathrm{2}{k}+\mathrm{1}\right)−{arctan}\left(\mathrm{2}{k}−\mathrm{1}\right)\right\}=\:−{arctan}\left(\mathrm{1}\right)+{arctan}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$

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