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Question Number 218196 by mnjuly1970 last updated on 01/Apr/25
lim _(n→∞) (1/n) ( (((2n)!)/(n!)) )^(1/n) = ?
$$ \\ $$$$ \\ $$$$\:\:\:\:\mathrm{lim}\:_{\mathrm{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\:\left(\:\frac{\left(\mathrm{2}{n}\right)!}{{n}!}\:\right)^{\frac{\mathrm{1}}{{n}}} =\:?\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by mehdee7396 last updated on 01/Apr/25
(1/n)((((2n)!)/(n!)))^(1/n) =((((2n)!)/(n^n n!)))^(1/n) =((((2n)(2n−1)(2n−2)...(n+2)(n+1))/(n×n×n×...×n×n)))^(1/n) =A lnA=(1/n)[ln2+ln(2−(1/n))+ln(2−(2/n))+...+ln(2−((n−1)/n))] (1/n)Σ_(k=1) ^n ln(2−((k−1)/n)) ⇒lim_(n→∞) (1/n)Σ_(k=1) ^n ln(2−((k−1)/n))=∫_0 ^1 ln(2−x)dx =(x−2)ln(2−x)−x]_0 ^1 =−1+2ln2=ln(4/e) ⇒A=(4/e) ✓
$$\frac{\mathrm{1}}{{n}}\left(\frac{\left(\mathrm{2}{n}\right)!}{{n}!}\right)^{\frac{\mathrm{1}}{{n}}} =\left(\frac{\left(\mathrm{2}{n}\right)!}{{n}^{{n}} {n}!}\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$=\left(\frac{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{2}\right)…\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)}{{n}×{n}×{n}×…×{n}×{n}}\right)^{\frac{\mathrm{1}}{{n}}} ={A} \\ $$$${lnA}=\frac{\mathrm{1}}{{n}}\left[{ln}\mathrm{2}+{ln}\left(\mathrm{2}−\frac{\mathrm{1}}{{n}}\right)+{ln}\left(\mathrm{2}−\frac{\mathrm{2}}{{n}}\right)+…+{ln}\left(\mathrm{2}−\frac{{n}−\mathrm{1}}{{n}}\right)\right] \\ $$$$\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{ln}\left(\mathrm{2}−\frac{{k}−\mathrm{1}}{{n}}\right) \\ $$$$\Rightarrow{li}\underset{{n}\rightarrow\infty} {{m}}\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{ln}\left(\mathrm{2}−\frac{{k}−\mathrm{1}}{{n}}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{2}−{x}\right){dx} \\ $$$$\left.=\left({x}−\mathrm{2}\right){ln}\left(\mathrm{2}−{x}\right)−{x}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=−\mathrm{1}+\mathrm{2}{ln}\mathrm{2}={ln}\frac{\mathrm{4}}{{e}} \\ $$$$\Rightarrow{A}=\frac{\mathrm{4}}{{e}}\:\checkmark \\ $$$$ \\ $$
Answered by vnm last updated on 01/Apr/25
using Stirling′s formula (1/n)((((2n)!)/(n!)))^(1/n) ∼_(n→∞) (1/n)((((√(2π∙2n))(((2n)/e))^(2n) )/( (√(2πn))((n/e))^n )))^(1/n) = (1/n)((√2))^(1/n) ∙(((((2n)/e))^2 )/(n/e))=((4((√2))^(1/n) )/e)→_(n→∞) (4/e)
$${using}\:{Stirling}'{s}\:{formula} \\ $$$$\frac{\mathrm{1}}{{n}}\left(\frac{\left(\mathrm{2}{n}\right)!}{{n}!}\right)^{\frac{\mathrm{1}}{{n}}} \underset{{n}\rightarrow\infty} {\sim}\frac{\mathrm{1}}{{n}}\left(\frac{\sqrt{\mathrm{2}\pi\centerdot\mathrm{2}{n}}\left(\frac{\mathrm{2}{n}}{{e}}\right)^{\mathrm{2}{n}} }{\:\sqrt{\mathrm{2}\pi{n}}\left(\frac{{n}}{{e}}\right)^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} = \\ $$$$\frac{\mathrm{1}}{{n}}\left(\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{n}}} \centerdot\frac{\left(\frac{\mathrm{2}{n}}{{e}}\right)^{\mathrm{2}} }{\frac{{n}}{{e}}}=\frac{\mathrm{4}\left(\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{n}}} }{{e}}\underset{{n}\rightarrow\infty} {\rightarrow}\frac{\mathrm{4}}{{e}} \\ $$
Commented by mnjuly1970 last updated on 01/Apr/25
$$\:\cancel{\underline{ }} \\ $$
Answered by vnm last updated on 02/Apr/25
without Stirling (((n!)^(1/n) )/n)→_(n→∞) e^(−1) (will be proved) (1/n)((((2n)!)/(n!)))^(1/n) =(1/n)(((2n∙(1/(2n))((2n)!)^(1/(2n)) )^2 )/(n∙(1/n)(n!)^(1/n) ))= ((4n^2 ((1/(2n))((2n)!)^(1/(2n)) )^2 )/(n^2 ∙(1/n)(n!)^(1/n) ))→_(n→∞) 4(((e^(−1) )^2 )/e^(−1) )=(4/e) ∫_2 ^(n+1) ln(x−1)∙dx<Σ_(k=1) ^n ln k<∫_1 ^(n+1) ln x∙dx (x−1)(ln (x−1)−1)∣_2 ^(n+1) <Σ_(k=1) ^n ln k<x(ln x−1)∣_1 ^(n+1) n(ln n−1)+1<Σ_(k=1) ^n ln k<(n+1)(ln(n+1)−1)+1 ln n−1+(1/n)<(1/n)Σ_(k=1) ^n ln k<((n+1)/n)ln(n+1)−1 ne^(−1+(1/n)) <((n!))^(1/n) <(n+1)^((n+1)/n) e^(−1) e^(−1+(1/n)) <(((n!))^(1/n) /n)<((n+1)/n)(n+1)^(1/n) e^(−1) lim_(n→∞) e^(1/n) =lim_(n→∞) ((n+1)/n)(n+1)^(1/n) =1⇒(((n!))^(1/n) /n)→_(n→∞) e^(−1)
$$ \\ $$$${without}\:{Stirling} \\ $$$$\frac{\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} }{{n}}\underset{{n}\rightarrow\infty} {\rightarrow}{e}^{−\mathrm{1}} \:\left({will}\:{be}\:{proved}\right) \\ $$$$\frac{\mathrm{1}}{{n}}\left(\frac{\left(\mathrm{2}{n}\right)!}{{n}!}\right)^{\frac{\mathrm{1}}{{n}}} =\frac{\mathrm{1}}{{n}}\frac{\left(\mathrm{2}{n}\centerdot\frac{\mathrm{1}}{\mathrm{2}{n}}\left(\left(\mathrm{2}{n}\right)!\right)^{\frac{\mathrm{1}}{\mathrm{2}{n}}} \right)^{\mathrm{2}} }{{n}\centerdot\frac{\mathrm{1}}{{n}}\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} }= \\ $$$$\frac{\mathrm{4}{n}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}{n}}\left(\left(\mathrm{2}{n}\right)!\right)^{\frac{\mathrm{1}}{\mathrm{2}{n}}} \right)^{\mathrm{2}} }{{n}^{\mathrm{2}} \centerdot\frac{\mathrm{1}}{{n}}\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} }\underset{{n}\rightarrow\infty} {\rightarrow}\mathrm{4}\frac{\left({e}^{−\mathrm{1}} \right)^{\mathrm{2}} }{{e}^{−\mathrm{1}} }=\frac{\mathrm{4}}{{e}} \\ $$$$ \\ $$$$\underset{\mathrm{2}} {\overset{{n}+\mathrm{1}} {\int}}\mathrm{ln}\left({x}−\mathrm{1}\right)\centerdot{dx}<\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{ln}\:{k}<\underset{\mathrm{1}} {\overset{{n}+\mathrm{1}} {\int}}\mathrm{ln}\:{x}\centerdot{dx} \\ $$$$\left({x}−\mathrm{1}\right)\left(\mathrm{ln}\:\left({x}−\mathrm{1}\right)−\mathrm{1}\right)\mid_{\mathrm{2}} ^{{n}+\mathrm{1}} <\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{ln}\:{k}<{x}\left(\mathrm{ln}\:{x}−\mathrm{1}\right)\mid_{\mathrm{1}} ^{{n}+\mathrm{1}} \\ $$$${n}\left(\mathrm{ln}\:{n}−\mathrm{1}\right)+\mathrm{1}<\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{ln}\:{k}<\left({n}+\mathrm{1}\right)\left(\mathrm{ln}\left({n}+\mathrm{1}\right)−\mathrm{1}\right)+\mathrm{1} \\ $$$$\mathrm{ln}\:{n}−\mathrm{1}+\frac{\mathrm{1}}{{n}}<\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{ln}\:{k}<\frac{{n}+\mathrm{1}}{{n}}\mathrm{ln}\left({n}+\mathrm{1}\right)−\mathrm{1} \\ $$$${ne}^{−\mathrm{1}+\frac{\mathrm{1}}{{n}}} <\sqrt[{{n}}]{{n}!}<\left({n}+\mathrm{1}\right)^{\frac{{n}+\mathrm{1}}{{n}}} {e}^{−\mathrm{1}} \\ $$$${e}^{−\mathrm{1}+\frac{\mathrm{1}}{{n}}} <\frac{\sqrt[{{n}}]{{n}!}}{{n}}<\frac{{n}+\mathrm{1}}{{n}}\left({n}+\mathrm{1}\right)^{\frac{\mathrm{1}}{{n}}} {e}^{−\mathrm{1}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{e}^{\frac{\mathrm{1}}{{n}}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}+\mathrm{1}}{{n}}\left({n}+\mathrm{1}\right)^{\frac{\mathrm{1}}{{n}}} =\mathrm{1}\Rightarrow\frac{\sqrt[{{n}}]{{n}!}}{{n}}\underset{{n}\rightarrow\infty} {\rightarrow}{e}^{−\mathrm{1}} \\ $$

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