Question-183936

Question Number 183936 by universe last updated on 31/Dec/22

Answered by Ar Brandon last updated on 31/Dec/22

L=lim_(n→∞) (Σ_(k=1) ^n (1/( ^n C_k )))^n =lim_(n→∞) (Σ_(k=1) ^n (((n−k)!k!)/(n!)))^n =lim_(n→∞) (Σ_(k=1) ^n ((Γ(n−k+1)Γ(k+1))/(Γ(n+1))))^n =lim_(n→∞) (Σ_(k=1) ^n (n+1)β(n−k+1, k+1))^n =lim_(n→∞) (Σ_(k=1) ^n (n+1)∫_0 ^1 x^(n−k) (1−x)^k dx)^n =lim_(n→∞) (∫_0 ^1 (n+1)x^n Σ_(k=1) ^n (((1−x)/x))^k dx)^n =lim_(n→∞) (∫_0 ^1 (n+1)x^n (((((1−x)/x))((((1−x)/x))^n −1))/((((1−x)/x))−1))dx)^n =lim_(n→∞) ((n+1)∫_0 ^1 (((1−x)((1−x)^n −x^n ))/(1−2x))dx)^n

$$\mathscr{L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\overset{{n}} {\:}{C}_{{k}} }\right)^{{n}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left({n}−{k}\right)!{k}!}{{n}!}\right)^{{n}} \\ $$$$\:\:\:\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\Gamma\left({n}−{k}+\mathrm{1}\right)\Gamma\left({k}+\mathrm{1}\right)}{\Gamma\left({n}+\mathrm{1}\right)}\right)^{{n}} \\ $$$$\:\:\:\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({n}+\mathrm{1}\right)\beta\left({n}−{k}+\mathrm{1},\:{k}+\mathrm{1}\right)\right)^{{n}} \\ $$$$\:\:\:\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({n}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−{k}} \left(\mathrm{1}−{x}\right)^{{k}} {dx}\right)^{{n}} \\ $$$$\:\:\:\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \left({n}+\mathrm{1}\right){x}^{{n}} \underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}−{x}}{{x}}\right)^{{k}} {dx}\right)^{{n}} \\ $$$$\:\:\:\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \left({n}+\mathrm{1}\right){x}^{{n}} \frac{\left(\frac{\mathrm{1}−{x}}{{x}}\right)\left(\left(\frac{\mathrm{1}−{x}}{{x}}\right)^{{n}} −\mathrm{1}\right)}{\left(\frac{\mathrm{1}−{x}}{{x}}\right)−\mathrm{1}}{dx}\right)^{{n}} \\ $$$$\:\:\:\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\left({n}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{x}\right)\left(\left(\mathrm{1}−{x}\right)^{{n}} −{x}^{{n}} \right)}{\mathrm{1}−\mathrm{2}{x}}{dx}\right)^{{n}} \\ $$

Commented by universe last updated on 01/Jan/23

$${Answer}\:?\:? \\ $$

Commented by Ar Brandon last updated on 01/Jan/23

Stucked ! sorry.

Commented by universe last updated on 01/Jan/23

$${no}\:{problem}\:{sir} \\ $$

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