Question Number 123826 by pticantor last updated on 28/Nov/20
$$ \\ $$$$\boldsymbol{{W}}_{\boldsymbol{{n}}} =\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\mathrm{2}\boldsymbol{{n}}} {\sum}}\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} } \\ $$$$\boldsymbol{{montrer}}\:\boldsymbol{{que}}\:\boldsymbol{{W}}_{\boldsymbol{{n}}} \:\boldsymbol{{converge}} \\ $$$$\boldsymbol{{et}}\:\boldsymbol{{calculer}}\:\boldsymbol{{la}}\:\boldsymbol{{valeur}}\:\boldsymbol{{de}}\:\boldsymbol{{W}}_{\boldsymbol{{n}}} \\ $$
Commented by Dwaipayan Shikari last updated on 28/Nov/20
![W_n =lim_(n→∞) Σ_(k=1) ^(2n) (k/(n^2 +k^2 )) (1/n)Σ_(k=1) ^(2n) ((k/n)/(1+((k/n))^2 ))=∫_0 ^2 (x/(1+x^2 ))dx =(1/2)[log(1+x^2 )]_0 ^2 =log((√5))](https://www.tinkutara.com/question/Q123830.png)
$$\boldsymbol{{W}}_{\boldsymbol{{n}}} =\underset{\boldsymbol{{n}}\rightarrow\infty} {\mathrm{lim}}\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\mathrm{2}\boldsymbol{{n}}} {\sum}}\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\boldsymbol{{n}}}\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\frac{{k}}{{n}}}{\mathrm{1}+\left(\frac{{k}}{{n}}\right)^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\mathrm{2}} \frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\left[{log}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{2}} ={log}\left(\sqrt{\mathrm{5}}\right) \\ $$
Answered by mathmax by abdo last updated on 28/Nov/20
![W_n =Σ_(k=1) ^n (k/(n^2 +k^2 )) +Σ_(k=n+1) ^(2n) (k/(n^2 +k^2 )) =u_n +v_n u_n =Σ_(k=1) ^n ((k/n^2 )/(1+(k^2 /n^2 ))) =(1/n)Σ_(k=1) ^n ((k/n)/(1+((k/n))^2 ))→∫_0 ^1 (x/(1+x^2 ))dx =(1/2)∫_0 ^1 ((2x)/(1+x^2 ))dx =(1/2)[ln(1+x^2 )]_0 ^1 =((ln(2))/2) v_n =Σ_(k=n+1) ^(2n) (k/(n^2 +k^2 )) =_(k−n=p) Σ_(p=1) ^n ((n+p)/(n^2 +(n+p)^2 )) =Σ_(p=1) ^n (((n+p)/n^2 )/(1+(((n+p)/n))^2 )) =(1/n)Σ_(p=1) ^n ((1+(p/n))/(1+(1+(p/n))^2 )) →∫_0 ^1 ((1+x)/(1+(1+x)^2 ))dx =_(1+x=t) ∫_1 ^2 (t/(1+t^2 ))dt =(1/2)[ln(1+t^2 )]_1 ^2 =(1/2){ln(5)−ln(2)} ⇒lim_(n→+∞) W_n =((ln2)/2) +((ln(5))/2)−((ln(2))/2) ⇒ lim_(n→+∞) =(1/2)ln(5)](https://www.tinkutara.com/question/Q123854.png)
$$\mathrm{W}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{k}}{\mathrm{n}^{\mathrm{2}} \:+\mathrm{k}^{\mathrm{2}} }\:+\sum_{\mathrm{k}=\mathrm{n}+\mathrm{1}} ^{\mathrm{2n}} \:\frac{\mathrm{k}}{\mathrm{n}^{\mathrm{2}} \:+\mathrm{k}^{\mathrm{2}} }\:=\mathrm{u}_{\mathrm{n}} +\mathrm{v}_{\mathrm{n}} \\ $$$$\mathrm{u}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\frac{\mathrm{k}}{\mathrm{n}^{\mathrm{2}} }}{\mathrm{1}+\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} }}\:=\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\frac{\mathrm{k}}{\mathrm{n}}}{\mathrm{1}+\left(\frac{\mathrm{k}}{\mathrm{n}}\right)^{\mathrm{2}} }\rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\mathrm{v}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{n}+\mathrm{1}} ^{\mathrm{2n}} \:\frac{\mathrm{k}}{\mathrm{n}^{\mathrm{2}} \:+\mathrm{k}^{\mathrm{2}} }\:=_{\mathrm{k}−\mathrm{n}=\mathrm{p}} \:\:\sum_{\mathrm{p}=\mathrm{1}} ^{\mathrm{n}} \:\:\frac{\mathrm{n}+\mathrm{p}}{\mathrm{n}^{\mathrm{2}} +\left(\mathrm{n}+\mathrm{p}\right)^{\mathrm{2}} } \\ $$$$=\sum_{\mathrm{p}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\frac{\mathrm{n}+\mathrm{p}}{\mathrm{n}^{\mathrm{2}} }}{\mathrm{1}+\left(\frac{\mathrm{n}+\mathrm{p}}{\mathrm{n}}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{p}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}+\frac{\mathrm{p}}{\mathrm{n}}}{\mathrm{1}+\left(\mathrm{1}+\frac{\mathrm{p}}{\mathrm{n}}\right)^{\mathrm{2}} } \\ $$$$\rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}+\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\mathrm{dx}\:\:=_{\mathrm{1}+\mathrm{x}=\mathrm{t}} \:\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{ln}\left(\mathrm{5}\right)−\mathrm{ln}\left(\mathrm{2}\right)\right\}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{W}_{\mathrm{n}} =\frac{\mathrm{ln2}}{\mathrm{2}}\:+\frac{\mathrm{ln}\left(\mathrm{5}\right)}{\mathrm{2}}−\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{5}\right) \\ $$