Question Number 164196 by HongKing last updated on 15/Jan/22
$$\mathrm{Evalute}\:\mathrm{the}\:\mathrm{sum}: \\ $$$$\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\mathrm{k}\:\left(\frac{\pi}{\mathrm{n}}\right)^{\mathrm{2}} \mathrm{arctan}\:\left(\frac{\mathrm{k}\pi}{\mathrm{n}}\right)^{\mathrm{2}} \\ $$
Answered by mathmax by abdo last updated on 15/Jan/22
![S_n =(π/n)Σ_(k=1) ^n (((kπ)/n))arctan(((kπ)/n))^2 ⇒lim_(n→+∞) S_n =∫_0 ^π xarctan(x^2 )dx =_(by parts) [(x^2 /2)arctan(x^2 )]_0 ^π −∫_0 ^π (x^2 /2)((2x)/(1+x^4 ))dx =(π^2 /2)arctan(π^2 )−∫_0 ^π (x^3 /(1+x^4 ))dx and ∫_0 ^(π ) (x^3 /(1+x^4 ))dx=[(1/(4 ))ln(1+x^4 )]_0 ^π =(1/4)ln(1+π^4 ) ⇒ lim_(n→+∞) S_n =(π^2 /2)arctan(π^2 )−(1/4)ln(1+π^4 )](https://www.tinkutara.com/question/Q164271.png)
$$\mathrm{S}_{\mathrm{n}} =\frac{\pi}{\mathrm{n}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\frac{\mathrm{k}\pi}{\mathrm{n}}\right)\mathrm{arctan}\left(\frac{\mathrm{k}\pi}{\mathrm{n}}\right)^{\mathrm{2}} \Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{S}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\pi} \mathrm{xarctan}\left(\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$=_{\mathrm{by}\:\mathrm{parts}} \left[\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{arctan}\left(\mathrm{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\pi} −\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\frac{\mathrm{2x}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\mathrm{arctan}\left(\pi^{\mathrm{2}} \right)−\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx}\:\:\mathrm{and} \\ $$$$\int_{\mathrm{0}} ^{\pi\:} \:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx}=\left[\frac{\mathrm{1}}{\mathrm{4}\:}\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} \right)\right]_{\mathrm{0}} ^{\pi} \:=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{1}+\pi^{\mathrm{4}} \right)\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{S}_{\mathrm{n}} =\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\mathrm{arctan}\left(\pi^{\mathrm{2}} \right)−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{1}+\pi^{\mathrm{4}} \right) \\ $$
Commented by HongKing last updated on 15/Jan/22
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir} \\ $$