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Question Number 158759 by HongKing last updated on 08/Nov/21

Prove that: lim_(n→∞) ((Σ_(k=0) ^(2n) (-1)^k ∙ ((4n + 1)/(4n - 2k + 1)) (((2n)),(( k)) )))^(1/n) = 1

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\boldsymbol{\mathrm{n}}}]{\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\overset{\mathrm{2}\boldsymbol{\mathrm{n}}} {\sum}}\left(-\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}} \:\centerdot\:\frac{\mathrm{4n}\:+\:\mathrm{1}}{\mathrm{4n}\:-\:\mathrm{2k}\:+\:\mathrm{1}}\begin{pmatrix}{\mathrm{2n}}\\{\:\mathrm{k}}\end{pmatrix}}\:=\:\mathrm{1} \\ $$$$ \\ $$

Answered by mindispower last updated on 08/Nov/21

Σ_(k≥0) ^(2n) (−1)^k .((4n+1)/(4n−2k+1)) (((2n)),(( k)) ) =Σ_(k=0) ^(2n) (−1)^k ((4n+1)/(2k+1)) (((2n)),(k) )=(4n+1)Σ_(k=0) ^(2n) ∫_0 ^1 .(−x^2 )^k (((2n)),(( k)) )dx =(4n+1)∫_0 ^1 (−x^2 )^k (((2n)),(k) )dx=(4n+1)∫^1 _0 (1−x^2 )^(2n) dx =(4n+1)∫_0 ^1 (1−t)^(2n) .t^(−(1/2)) (dt/2) =(2n+(1/2))β(2n+1,(1/2))=((Γ(2n+1)Γ((1/2)))/(Γ(2n+1+(1/2)))) =((Γ(2n+1)Γ((1/2)))/(Γ(2n+(1/2)))) lim_(n→∞) .(((Γ((1/2))Γ(2n+1))/(Γ(2n+(1/2))))^(1/2) Γ(2n+(1/2))∼(√(2π)).(2n)^(2n) e^(−2n) Γ(2n+1)∼(√(2π))(2n+1)^(2n+(1/2)) e^(−2n) ⇔lim_(n→∞) ((((√π). (2n+1)^(2n+(1/2)) )/((2n)^(2n) )))^(1/n) =lim_(n→∞) .π^(1/(2n)) .(((2n+1)/(2n)))^2 .(2n+1)^(1/(2n)) =lim_(n→∞) π^0 .(1)^2 .e^((ln(1+2n))/(2n)) =1.1.1=1

$$\underset{{k}\geqslant\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} .\frac{\mathrm{4}{n}+\mathrm{1}}{\mathrm{4}{n}−\mathrm{2}{k}+\mathrm{1}}\begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{k}}\end{pmatrix} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{4}{n}+\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\begin{pmatrix}{\mathrm{2}{n}}\\{{k}}\end{pmatrix}=\left(\mathrm{4}{n}+\mathrm{1}\right)\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} .\left(−{x}^{\mathrm{2}} \right)^{{k}} \begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{k}}\end{pmatrix}{dx} \\ $$$$=\left(\mathrm{4}{n}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \left(−{x}^{\mathrm{2}} \right)^{{k}} \begin{pmatrix}{\mathrm{2}{n}}\\{{k}}\end{pmatrix}{dx}=\left(\mathrm{4}{n}+\mathrm{1}\right)\underset{\mathrm{0}} {\int}^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}{n}} {dx} \\ $$$$=\left(\mathrm{4}{n}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\mathrm{2}{n}} .{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \frac{{dt}}{\mathrm{2}} \\ $$$$=\left(\mathrm{2}{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\beta\left(\mathrm{2}{n}+\mathrm{1},\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\Gamma\left(\mathrm{2}{n}+\mathrm{1}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{2}{n}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\frac{\Gamma\left(\mathrm{2}{n}+\mathrm{1}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{2}{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}.\left(\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\mathrm{2}{n}+\mathrm{1}\right)}{\Gamma\left(\mathrm{2}{n}+\frac{\mathrm{1}}{\mathrm{2}}\right.}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Gamma\left(\mathrm{2}{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sim\sqrt{\mathrm{2}\pi}.\left(\mathrm{2}{n}\right)^{\mathrm{2}{n}} {e}^{−\mathrm{2}{n}} \\ $$$$\Gamma\left(\mathrm{2}{n}+\mathrm{1}\right)\sim\sqrt{\mathrm{2}\pi}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}{n}+\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−\mathrm{2}{n}} \\ $$$$\Leftrightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\sqrt{\pi}.\:\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}{n}+\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{2}{n}\right)^{\mathrm{2}{n}} }\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}.\pi^{\frac{\mathrm{1}}{\mathrm{2}{n}}} .\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}}\right)^{\mathrm{2}} .\left(\mathrm{2}{n}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}{n}}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\pi^{\mathrm{0}} .\left(\mathrm{1}\right)^{\mathrm{2}} .{e}^{\frac{{ln}\left(\mathrm{1}+\mathrm{2}{n}\right)}{\mathrm{2}{n}}} =\mathrm{1}.\mathrm{1}.\mathrm{1}=\mathrm{1} \\ $$$$ \\ $$

Commented by HongKing last updated on 09/Nov/21

thank you so much my dear Ser perfect

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{perfect} \\ $$

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