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Question Number 107288 by mathmax by abdo last updated on 09/Aug/20
$$\mathrm{let}\:\mathrm{u}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\mathrm{n}^{\mathrm{2}} \:+\mathrm{k}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{n}}} \\ $$$$\mathrm{determine}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{u}_{\mathrm{n}} \\ $$
Answered by Ar Brandon last updated on 13/Aug/20
![u_n =(1/n^2 )Π_(k=1) ^n (n^2 +k^2 )^(1/n) ln(u_n )=−2ln(n)+lnΠ_(k=1) ^n (n^2 +k^2 )^(1/n) =−2ln(n)+(1/n)Σ_(k=1) ^n ln(n^2 +k^2 ) =−2ln(n)+(1/n)Σ_(k=1) ^n [2ln(n)+ln(1+(k^2 /n^2 ))] =−2ln(n)+2ln(n)+(1/n)Σ_(k=1) ^n ln(1+(k^2 /n^2 )) lim_(n→∞) ln(u_n )=lim_(n→∞) (1/n)Σ_(k=1) ^n ln(1+(k^2 /n^2 )) =∫_0 ^1 ln(1+x^2 )dx =[ln(1+x^2 )∫dx−∫{((d ln(1+x^2 ))/dx)∙∫dx}dx]_0 ^1 =[xln(1+x^2 )−∫((2x^2 )/(1+x^2 ))dx]_0 ^1 =ln2−2∫_0 ^1 {1−(1/(1+x^2 ))}dx =ln2−2+(π/2) ⇒lim_(n→∞) u_n =e^(ln2−2+(π/2)) =2e^((π/2)−2)](Q107874.png)
$$\mathrm{u}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{n}^{\mathrm{2}} +\mathrm{k}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{n}}} \\ $$$$\mathrm{ln}\left(\mathrm{u}_{\mathrm{n}} \right)=−\mathrm{2ln}\left(\mathrm{n}\right)+\mathrm{ln}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{n}^{\mathrm{2}} +\mathrm{k}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{2ln}\left(\mathrm{n}\right)+\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{ln}\left(\mathrm{n}^{\mathrm{2}} +\mathrm{k}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{2ln}\left(\mathrm{n}\right)+\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left[\mathrm{2ln}\left(\mathrm{n}\right)+\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} }\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{2ln}\left(\mathrm{n}\right)+\mathrm{2ln}\left(\mathrm{n}\right)+\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} }\right) \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}ln}\left(\mathrm{u}_{\mathrm{n}} \right)=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\int\mathrm{dx}−\int\left\{\frac{\mathrm{d}\:\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{dx}}\centerdot\int\mathrm{dx}\right\}\mathrm{dx}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[\mathrm{xln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)−\int\frac{\mathrm{2x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{ln2}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right\}\mathrm{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{ln2}−\mathrm{2}+\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}u}_{\mathrm{n}} =\mathrm{e}^{\mathrm{ln2}−\mathrm{2}+\frac{\pi}{\mathrm{2}}} =\mathrm{2e}^{\frac{\pi}{\mathrm{2}}−\mathrm{2}} \\ $$