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

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

$$\mathcal{L}=\underset{n\to \mathrm{\infty }}{\lim }{\left(\underset{k=1}{\sum ^n}\frac{1}{\stackrel{n}{}{C}_k}\right)}^n=\underset{n\to \mathrm{\infty }}{\lim }{\left(\underset{k=1}{\sum ^n}\frac{(n-k)!k!}{n!}\right)}^n$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\left(\underset{k=1}{\sum ^n}\frac{\mathrm{\Gamma }(n-k+1)\mathrm{\Gamma }(k+1)}{\mathrm{\Gamma }(n+1)}\right)}^n$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\left(\underset{k=1}{\sum ^n}(n+1)\beta (n-k+1,k+1)\right)}^n$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\left(\underset{k=1}{\sum ^n}(n+1){\int }_0^1{x}^{n-k}{(1-x)}^{k}dx\right)}^n$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\left({\int }_0^1(n+1)x^n\underset{k=1}{\sum ^n}{\left(\frac{1-x}{x}\right)}^{k}dx\right)}^n$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\left({\int }_0^1(n+1)x^n\frac{\left(\frac{1-x}{x}\right)({\left(\frac{1-x}{x}\right)}^n-1)}{\left(\frac{1-x}{x}\right)-1}dx\right)}^n$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\left((n+1){\int }_0^1\frac{(1-x)({(1-x)}^n-x^n)}{1-2x}dx\right)}^n$$

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

$$noproblemsir$$

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