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Question Number 195393 by Erico last updated on 01/Aug/23
prove that lim_(x→0) (((Σ_(k=1) ^n (1−(1/(2k)))^x )/n))^(1/( x )) = (1/4)(C_(2n) ^n )^(1/n)
$$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt[{\:\:\boldsymbol{{x}}\:\:}]{\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{k}}\right)^{{x}} }{{n}}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\sqrt[{\boldsymbol{{n}}}]{\mathrm{C}_{\mathrm{2}\boldsymbol{\mathrm{n}}} ^{\boldsymbol{\mathrm{n}}} } \\ $$
Answered by witcher3 last updated on 01/Aug/23
f(t)=t^x =e^(xln(t)) =1+xln(t)+o(x),x→0 Σ_(k=1) ^n (1−(1/(2k)))^x =Σ(1+xln(1−(1/(2k)))+o(x)) =n+xln(Π_(k=1) ^n (((2k−1)/(2k))))+o(x) =n+xln(Π_(k=1) ^n (((2k)(2k−1))/(4k^2 )))+o(x) =n+xln((((2n)!)/(4^n (n!)^2 )))+o(x)=n+xln((1/4^n )C_(2n) ^n )+o(x) lim_(x→0) (((1/n)Σ_(k=1) ^n (1−(1/(2k)))^x ))^(1/x) =lim_(x→0) e^((1/x)ln((1/n)(n+xln((1/4^n )C_(2n) ^n )+o(x))) =lim_(x→0) e^((ln(1+(x/n)ln((1/4^n )C_(2n) ^n )+o(1)))/x) =e^(lim_(x→0) ((ln(1+(x/n)ln((C_(2n) ^n /4^n ))+o(1)))/x)) ≪exp is continus≫ =e^(lim_(x→0) (((x/n)ln((C_(2n) ^n /4^n ))+o(1))/x)) =e^(ln(((C_(2n) ^n /4^n ))^(1/n) )) =((C_(2n) ^n /4^n ))^(1/n) (1/4)(C_(2n) ^n )^(1/n)
$$\mathrm{f}\left(\mathrm{t}\right)=\mathrm{t}^{\mathrm{x}} =\mathrm{e}^{\mathrm{xln}\left(\mathrm{t}\right)} =\mathrm{1}+\mathrm{xln}\left(\mathrm{t}\right)+\mathrm{o}\left(\mathrm{x}\right),\mathrm{x}\rightarrow\mathrm{0} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2k}}\right)^{\mathrm{x}} =\Sigma\left(\mathrm{1}+\mathrm{xln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2k}}\right)+\mathrm{o}\left(\mathrm{x}\right)\right) \\ $$$$=\mathrm{n}+\mathrm{xln}\left(\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\frac{\mathrm{2k}−\mathrm{1}}{\mathrm{2k}}\right)\right)+\mathrm{o}\left(\mathrm{x}\right) \\ $$$$=\mathrm{n}+\mathrm{xln}\left(\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\frac{\left(\mathrm{2k}\right)\left(\mathrm{2k}−\mathrm{1}\right)}{\mathrm{4k}^{\mathrm{2}} }\right)+\mathrm{o}\left(\mathrm{x}\right) \\ $$$$=\mathrm{n}+\mathrm{xln}\left(\frac{\left(\mathrm{2n}\right)!}{\mathrm{4}^{\mathrm{n}} \left(\mathrm{n}!\right)^{\mathrm{2}} }\right)+\mathrm{o}\left(\mathrm{x}\right)=\mathrm{n}+\mathrm{xln}\left(\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{n}} }\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \right)+\mathrm{o}\left(\mathrm{x}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\sqrt[{\mathrm{x}}]{\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2k}}\right)^{\mathrm{x}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}e}^{\frac{\mathrm{1}}{\mathrm{x}}\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{n}}\left(\mathrm{n}+\mathrm{xln}\left(\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{n}} }\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \right)+\mathrm{o}\left(\mathrm{x}\right)\right)\right.} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}e}^{\frac{\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{n}}\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{n}} }\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \right)+\mathrm{o}\left(\mathrm{1}\right)\right)}{\mathrm{x}}} =\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{n}}\mathrm{ln}\left(\frac{\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} }{\mathrm{4}^{\mathrm{n}} }\right)+\mathrm{o}\left(\mathrm{1}\right)\right)}{\mathrm{x}}} \\ $$$$\ll\mathrm{exp}\:\mathrm{is}\:\mathrm{continus}\gg \\ $$$$=\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{x}}{\mathrm{n}}\mathrm{ln}\left(\frac{\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} }{\mathrm{4}^{\mathrm{n}} }\right)+\mathrm{o}\left(\mathrm{1}\right)}{\mathrm{x}}} =\mathrm{e}^{\mathrm{ln}\left(\sqrt[{\mathrm{n}}]{\frac{\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} }{\mathrm{4}^{\mathrm{n}} }}\right)} =\sqrt[{\mathrm{n}}]{\frac{\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} }{\mathrm{4}^{\mathrm{n}} }} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\sqrt[{\mathrm{n}}]{\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} } \\ $$

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