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Question Number 204477 by universe last updated on 18/Feb/24
lim_(n→∞) (2n∫_0 ^1 (x^n /(1+x^2 ))dx)^n =?
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2n}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\right)^{\mathrm{n}} =? \\ $$
Answered by witcher3 last updated on 18/Feb/24
∫_0 ^1 (x^n /((1+x^2 )))dx=∫_0 ^1 ((x^n −x^(n+2) )/(1−x^4 ))dx =(1/4)∫_0 ^1 ((t^((n/4)−(3/4)) −t^(((n+2)/4)−(3/4)) )/(1−t))dt=(1/4)(𝚿((n/4)+(3/4))−Ψ((n/4)+(1/4)))=f(n) Ψ(z)=ln(z)−(1/(2z))−(1/(12(z^2 )))+o((1/z^3 )) Ψ(((n+3)/4))−Ψ(((n+1)/4))=ln(((n+3)/(n+1)))−(1/2)((4/(n+3))−(4/(n+1)))−(1/(12))(((16)/((n+3)^2 ))−((16)/((n+1)^2 )))+o((1/n^3 )) ln(1+(3/n))−ln(1+(1/n))−(2/n)((1/(1+(3/n)))−(1/(1+(1/n))))−(4/(3n^2 ))((1/((1+(3/n))^2 ))−(1/((1+(1/n))^2 ))) (1/((1+x)^2 ))=Σ_(k≥1) (−1)^(k−1) kx^(k−1) =(2/n)−(4/n^2 )−(2/n)(1−(3/n)+(9/n^2 )−(1−(1/n)+(1/n^2 )))−(4/(3n^2 ))(1−(6/n)−(1−(2/n)))+o((1/n^3 )) =(2/n)−((16)/n^3 )+((16)/(3n^3 ))+o((1/n^3 )) =(2/n)−((32)/(3n^3 ))+o((1/n^3 )) (n/2)((2/n)−((32)/(3n^3 ))+o((1/n^3 )))=1−((16)/(3n^2 ))+o((1/n^2 )) e^(nln(1−((16)/(3n^2 ))+o((1/n^2 )))) =e^(−((16)/(3n))+o((1/n))) →1
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{n}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{n}} −\mathrm{x}^{\mathrm{n}+\mathrm{2}} }{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{t}^{\frac{\mathrm{n}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{4}}} −\boldsymbol{\mathrm{t}}^{\frac{\boldsymbol{\mathrm{n}}+\mathrm{2}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{4}}} }{\mathrm{1}−\mathrm{t}}\boldsymbol{\mathrm{dt}}=\frac{\mathrm{1}}{\mathrm{4}}\left(\boldsymbol{\Psi}\left(\frac{\mathrm{n}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}\right)−\Psi\left(\frac{\mathrm{n}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\right)\right)=\mathrm{f}\left(\mathrm{n}\right) \\ $$$$\Psi\left(\mathrm{z}\right)=\mathrm{ln}\left(\mathrm{z}\right)−\frac{\mathrm{1}}{\mathrm{2z}}−\frac{\mathrm{1}}{\mathrm{12}\left(\mathrm{z}^{\mathrm{2}} \right)}+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{3}} }\right) \\ $$$$\Psi\left(\frac{\mathrm{n}+\mathrm{3}}{\mathrm{4}}\right)−\Psi\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{4}}\right)=\mathrm{ln}\left(\frac{\mathrm{n}+\mathrm{3}}{\mathrm{n}+\mathrm{1}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{4}}{\mathrm{n}+\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{n}+\mathrm{1}}\right)−\frac{\mathrm{1}}{\mathrm{12}}\left(\frac{\mathrm{16}}{\left(\mathrm{n}+\mathrm{3}\right)^{\mathrm{2}} }−\frac{\mathrm{16}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\right) \\ $$$$\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{n}}\right)−\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)−\frac{\mathrm{2}}{\mathrm{n}}\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{n}}}−\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}\right)−\frac{\mathrm{4}}{\mathrm{3n}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{n}}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }=\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \mathrm{kx}^{\mathrm{k}−\mathrm{1}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{n}}−\frac{\mathrm{4}}{\mathrm{n}^{\mathrm{2}} }−\frac{\mathrm{2}}{\mathrm{n}}\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{n}}+\frac{\mathrm{9}}{\mathrm{n}^{\mathrm{2}} }−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)\right)−\frac{\mathrm{4}}{\mathrm{3n}^{\mathrm{2}} }\left(\mathrm{1}−\frac{\mathrm{6}}{\mathrm{n}}−\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{n}}\right)\right)+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{n}}−\frac{\mathrm{16}}{\mathrm{n}^{\mathrm{3}} }+\frac{\mathrm{16}}{\mathrm{3n}^{\mathrm{3}} }+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{n}}−\frac{\mathrm{32}}{\mathrm{3n}^{\mathrm{3}} }+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\right) \\ $$$$\frac{\mathrm{n}}{\mathrm{2}}\left(\frac{\mathrm{2}}{\mathrm{n}}−\frac{\mathrm{32}}{\mathrm{3n}^{\mathrm{3}} }+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\right)\right)=\mathrm{1}−\frac{\mathrm{16}}{\mathrm{3n}^{\mathrm{2}} }+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right) \\ $$$$\mathrm{e}^{\mathrm{nln}\left(\mathrm{1}−\frac{\mathrm{16}}{\mathrm{3n}^{\mathrm{2}} }+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)\right)} =\mathrm{e}^{−\frac{\mathrm{16}}{\mathrm{3n}}+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}}\right)} \rightarrow\mathrm{1} \\ $$
Commented by universe last updated on 19/Feb/24
thank you so much sir
$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$
Commented by witcher3 last updated on 19/Feb/24
withe pleasur i think we can prove it withe high school tools
$$\mathrm{withe}\:\mathrm{pleasur}\:\mathrm{i}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{prove}\:\mathrm{it}\:\mathrm{withe}\:\mathrm{high}\:\mathrm{school}\:\mathrm{tools} \\ $$
Commented by universe last updated on 19/Feb/24
how sir can u prove it??
$${how}\:{sir}\:{can}\:{u}\:{prove}\:{it}?? \\ $$

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