# let-A-n-k-1-n-1-k-2-n-2-calculate-lim-n-ln-A-n-n-

Question Number 66172 by mathmax by abdo last updated on 10/Aug/19
$$\\$$$${let}\:{A}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}+\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\:\:\:{calculate}\:{lim}_{{n}\rightarrow+\infty} \:\frac{{ln}\left({A}_{{n}} \right)}{{n}} \\$$
Commented by Prithwish sen last updated on 11/Aug/19
$$\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{n}}\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{ln}\left[\mathrm{1}+\left(\frac{\mathrm{k}}{\mathrm{n}}\right)^{\mathrm{2}} \right]\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx} \\$$$$=\left[\mathrm{xln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)−\mathrm{2x}+\mathrm{2tan}^{−\mathrm{1}} \mathrm{x}\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{ln2}\:−\mathrm{2}+\frac{\pi}{\mathrm{2}} \\$$$$\mathrm{please}\:\mathrm{check}. \\$$
Commented by mathmax by abdo last updated on 10/Aug/19
$${we}\:{have}\:{ln}\left({A}_{{n}} \right)=\sum_{{k}=\mathrm{1}} ^{{n}} {ln}\left(\mathrm{1}+\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\:\Rightarrow\frac{{ln}\left({A}_{{n}} \right)}{{n}}\:=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} {ln}\left(\mathrm{1}+\left(\frac{{k}}{{n}}\right)^{\mathrm{2}} \right) \\$$$${we}\:{get}\:{a}\:{Rieman}\:{sum}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \frac{{ln}\left({A}_{{n}} \right)}{{n}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dx} \\$$$${by}\:{parts}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dx}\:=\left[{xln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} {x}\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$$$={ln}\left(\mathrm{2}\right)−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:={ln}\left(\mathrm{2}\right)−\mathrm{2}\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\$$$$={ln}\left(\mathrm{2}\right)−\mathrm{2}\:+\mathrm{2}\left[{arctanx}\right]_{\mathrm{0}} ^{\mathrm{1}} ={ln}\left(\mathrm{2}\right)−\mathrm{2}+\mathrm{2}×\frac{\pi}{\mathrm{4}}\:={ln}\left(\mathrm{2}\right)−\mathrm{2}+\frac{\pi}{\mathrm{2}} \\$$$${finally}\:{lim}_{{n}\rightarrow+\infty} \:\frac{{ln}\left({A}_{{n}} \right)}{{n}}\:=\frac{\pi}{\mathrm{2}}\:+{ln}\left(\mathrm{2}\right)−\mathrm{2}\:. \\$$
Commented by Prithwish sen last updated on 10/Aug/19
$$\mathrm{thanks}\:\mathrm{sir}. \\$$
Commented by mathmax by abdo last updated on 10/Aug/19
$${you}\:{are}\:{welcome}\:{sir}. \\$$