Question Number 90751 by abdomathmax last updated on 25/Apr/20
$$\:{calculste}\:{lim}_{{n}\rightarrow\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\sum_{{k}=\mathrm{1}} ^{{n}} \:{karctan}\left(\frac{{k}}{{n}}\right) \\ $$
Commented by mathmax by abdo last updated on 26/Apr/20
![S_n =(1/n)Σ_(k=1) ^n (k/n) arctan((k/n)) ⇒ S_n is Rieman sum and lim_(n→+∞) S_n =∫_0 ^1 x arctan(x)dx by parts ∫_0 ^1 xarctan(x)dx =[(x^2 /2)arctan(x)]_0 ^1 −∫_0 ^1 (x^2 /2)×(dx/(1+x^2 )) =(π/8)−(1/2) ∫_0 ^1 ((1+x^2 −1)/(1+x^2 ))dx =(π/8)−(1/2) +(1/2)[arctanx]_0 ^1 =(π/8)−(1/2)+(π/8) =(π/4)−(1/2) ⇒lim_(n→+∞) S_n =(π/4)−(1/2)](https://www.tinkutara.com/question/Q90899.png)
$${S}_{{n}} =\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{k}}{{n}}\:{arctan}\left(\frac{{k}}{{n}}\right)\:\Rightarrow\:{S}_{{n}} {is}\:{Rieman}\:{sum}\:{and} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\:{arctan}\left({x}\right){dx}\:\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{xarctan}\left({x}\right){dx}\:=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{arctan}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\mathrm{2}}×\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\left[{arctanx}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\:=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$