Question Number 204991 by tigrecomplexe last updated on 04/Mar/24
Commented by tigrecomplexe last updated on 05/Mar/24
$${sommeone}\:{can}\:{put}\:{hand}\:{here}\:{please}\:? \\ $$
Answered by witcher3 last updated on 05/Mar/24
![u_n =lim_(n→∞) e^(sin(2π(√(1+n^2 )))Σ_(k=1) ^n ((ln(1+(k/n)))/2)) sin(2π(√(1+n^2 )))=sin(2πn(1+(1/n^2 ))^(1/2) ) (1+(1/n^2 ))^(1/2) =1+(1/(2n^2 ))+o((1/n^2 )) sin(2π(√(1+n^2 )))=^(n→∞) sin(2πn+(π/n)+o((1/n)))=sin((π/n)+o((1/n))) sin(2π(√(1+n^2 )))=^∞ (π/n)+o((1/n)) u_n =e^(nsin(2π(√(1+n^2 ))).(1/(2n))Σ_(k=1) ^n ln(1+(k/n))) lim_(n→∞) .(1/2).(1/n)Σ_(k=1) ^n ln(1+(k/n))=∫_0 ^1 ((ln(1+x))/2)=[(((1+x))/2)ln(1+x)−(x/2)]_0 ^1 =ln(2)−(1/2) lim_(n→∞) .nsin(2π(√(1+n^2 )))=lim_(n→∞) n.((π/n))=π U_n →e^(π(ln(2)−(1/2))) =((2π)/e^(π/2) )](https://www.tinkutara.com/question/Q205007.png)
$$\mathrm{u}_{\mathrm{n}} =\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}e}^{\mathrm{sin}\left(\mathrm{2}\pi\sqrt{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\right)\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}\right)}{\mathrm{2}}} \\ $$$$\mathrm{sin}\left(\mathrm{2}\pi\sqrt{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\right)=\mathrm{sin}\left(\mathrm{2}\pi\mathrm{n}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right) \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} }+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right) \\ $$$$\mathrm{sin}\left(\mathrm{2}\pi\sqrt{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\right)\overset{\mathrm{n}\rightarrow\infty} {=}\mathrm{sin}\left(\mathrm{2}\pi\mathrm{n}+\frac{\pi}{\mathrm{n}}+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}}\right)\right)=\mathrm{sin}\left(\frac{\pi}{\mathrm{n}}+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}}\right)\right) \\ $$$$\mathrm{sin}\left(\mathrm{2}\pi\sqrt{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\right)\overset{\infty} {=}\frac{\pi}{\mathrm{n}}+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}}\right) \\ $$$$\mathrm{u}_{\mathrm{n}} =\mathrm{e}^{\mathrm{nsin}\left(\mathrm{2}\pi\sqrt{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\right).\frac{\mathrm{1}}{\mathrm{2n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}\right)} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}.\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{2}}=\left[\frac{\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)−\frac{\mathrm{x}}{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}.\mathrm{nsin}\left(\mathrm{2}\pi\sqrt{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\right)=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{n}.\left(\frac{\pi}{\mathrm{n}}\right)=\pi \\ $$$$\mathrm{U}_{\mathrm{n}} \rightarrow\mathrm{e}^{\pi\left(\mathrm{ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\right)} =\frac{\mathrm{2}\pi}{\mathrm{e}^{\frac{\pi}{\mathrm{2}}} } \\ $$$$ \\ $$
Answered by Mathspace last updated on 05/Mar/24
![A_n =(Π_(k=1) ^n (√(1+(k/n))))^(sin(2π(√(1+n^2 )))) ⇒log(A_n )=sin(2π(√(1+n^2 )))Σ_(k=1) ^n log((√(1+(k/n)))) (√(1+n^2 ))=n(√(1+(1/n^2 )))=n(1+(1/n^2 ))^(1/2) ∼n(1+(1/(2n^2 )))=n+(1/(2n)) ⇒ sin(2π(√(1+n^2 )))∼sin(2πn+(π/n)) =sin((π/n))∼(π/n) ⇒ log(A_n )∼(π/n)Σ_(k=1) ^n log(√(1+(k/n))) Reiman sum give lim_(n→+∞) log(A_n ) =π∫_0 ^1 log(√(1+x))dx =(π/2)∫_0 ^1 log(1+x)dx but ∫_0 ^1 log(1+x)dx= [xlog(1+x)]_0 ^1 −∫_0 ^1 (x/(1+x))dx =log2−∫_0 ^1 ((1+x−1)/(1+x))dx =log2−1+∫_0 ^1 (dx/(1+x)) =log2−1+[log(1+x)]_0 ^1 =log2−1+log2=2log2−1 lim log(A_n )=(π/2)(2log2−1) =πlog2−(π/2) ⇒ lim_(n→+∞) A_n =e^(πlog2−(π/2)) =2^π ×e^(−(π/2))](https://www.tinkutara.com/question/Q205016.png)
$${A}_{{n}} =\left(\prod_{{k}=\mathrm{1}} ^{{n}} \sqrt{\mathrm{1}+\frac{{k}}{{n}}}\right)^{{sin}\left(\mathrm{2}\pi\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }\right)} \\ $$$$\Rightarrow{log}\left({A}_{{n}} \right)={sin}\left(\mathrm{2}\pi\sqrt{\left.\mathrm{1}+{n}^{\mathrm{2}} \right)}\sum_{{k}=\mathrm{1}} ^{{n}} {log}\left(\sqrt{\mathrm{1}+\frac{{k}}{{n}}}\right)\right. \\ $$$$\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }={n}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }}={n}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\sim{n}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }\right)={n}+\frac{\mathrm{1}}{\mathrm{2}{n}}\:\Rightarrow \\ $$$${sin}\left(\mathrm{2}\pi\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }\right)\sim{sin}\left(\mathrm{2}\pi{n}+\frac{\pi}{{n}}\right) \\ $$$$={sin}\left(\frac{\pi}{{n}}\right)\sim\frac{\pi}{{n}}\:\Rightarrow \\ $$$${log}\left({A}_{{n}} \right)\sim\frac{\pi}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} {log}\sqrt{\mathrm{1}+\frac{{k}}{{n}}} \\ $$$${Reiman}\:{sum}\:{give} \\ $$$${lim}_{{n}\rightarrow+\infty} {log}\left({A}_{{n}} \right) \\ $$$$=\pi\int_{\mathrm{0}} ^{\mathrm{1}} {log}\sqrt{\mathrm{1}+{x}}{dx} \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left(\mathrm{1}+{x}\right){dx}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left(\mathrm{1}+{x}\right){dx}= \\ $$$$\left[{xlog}\left(\mathrm{1}+{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\mathrm{1}+{x}}{dx} \\ $$$$={log}\mathrm{2}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}+{x}−\mathrm{1}}{\mathrm{1}+{x}}{dx} \\ $$$$={log}\mathrm{2}−\mathrm{1}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}} \\ $$$$={log}\mathrm{2}−\mathrm{1}+\left[{log}\left(\mathrm{1}+{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$={log}\mathrm{2}−\mathrm{1}+{log}\mathrm{2}=\mathrm{2}{log}\mathrm{2}−\mathrm{1} \\ $$$${lim}\:{log}\left({A}_{{n}} \right)=\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{log}\mathrm{2}−\mathrm{1}\right) \\ $$$$=\pi{log}\mathrm{2}−\frac{\pi}{\mathrm{2}}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} {A}_{{n}} ={e}^{\pi{log}\mathrm{2}−\frac{\pi}{\mathrm{2}}} \\ $$$$=\mathrm{2}^{\pi} ×{e}^{−\frac{\pi}{\mathrm{2}}} \\ $$$$ \\ $$