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Question Number 104016 by Dwaipayan Shikari last updated on 18/Jul/20

Answered by OlafThorendsen last updated on 18/Jul/20

arctanx−arctany = arctan((x−y)/(1+xy)) arctan(2k+1)−arctan(2k−1) = arctan(((2k+1)−(2k−1))/(1+(2k+1)(2k−1))) = arctan(2/(4k^2 )) = arctan(1/(2k^2 )) S_n =Σ_(k=1) ^(k=n) arctan(1/(2k^2 )) S_n =Σ_(k=1) ^(k=n) arctan(2k+1)−arctan(2k−1) S_n = arctan3−arctan1 +arctan5−arctan3... +arctan(2k+1)−arctan(2k−1) S_n = arctan(2k+1)−arctan1 lim_(n→∞) S_n = (π/2)−(π/4) = (π/4) Σ_(k=1) ^∞ arctan(1/(2k^2 )) = (π/4)

$$\mathrm{arctan}{x}−\mathrm{arctan}{y}\:=\:\mathrm{arctan}\frac{{x}−{y}}{\mathrm{1}+{xy}} \\ $$$$\mathrm{arctan}\left(\mathrm{2}{k}+\mathrm{1}\right)−\mathrm{arctan}\left(\mathrm{2}{k}−\mathrm{1}\right)\:=\:\mathrm{arctan}\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)−\left(\mathrm{2}{k}−\mathrm{1}\right)}{\mathrm{1}+\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}−\mathrm{1}\right)} \\ $$$$=\:\:\mathrm{arctan}\frac{\mathrm{2}}{\mathrm{4}{k}^{\mathrm{2}} }\:=\:\mathrm{arctan}\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} } \\ $$$$\mathrm{S}_{{n}} \:=\underset{{k}=\mathrm{1}} {\overset{{k}={n}} {\sum}}\mathrm{arctan}\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} } \\ $$$$\mathrm{S}_{{n}} \:=\underset{{k}=\mathrm{1}} {\overset{{k}={n}} {\sum}}\mathrm{arctan}\left(\mathrm{2}{k}+\mathrm{1}\right)−\mathrm{arctan}\left(\mathrm{2}{k}−\mathrm{1}\right) \\ $$$$\mathrm{S}_{{n}} \:=\:\mathrm{arctan3}−\mathrm{arctan1} \\ $$$$+\mathrm{arctan5}−\mathrm{arctan3}... \\ $$$$+\mathrm{arctan}\left(\mathrm{2}{k}+\mathrm{1}\right)−\mathrm{arctan}\left(\mathrm{2}{k}−\mathrm{1}\right) \\ $$$$\mathrm{S}_{{n}} \:=\:\mathrm{arctan}\left(\mathrm{2}{k}+\mathrm{1}\right)−\mathrm{arctan1} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}S}_{{n}} \:=\:\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\:=\:\frac{\pi}{\mathrm{4}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{arctan}\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} }\:=\:\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$

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