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Question Number 205885 by hardmath last updated on 01/Apr/24

$$\mathrm{Find}:\underset{\mathbf{n}\to \mathrm{\infty }}{\lim }\sqrt[\mathbf{n}]{}\left(\begin{array}{c}2\mathrm{n}\\ \mathrm{n}\end{array}\right)=?$$

Answered by MM42 last updated on 03/Apr/24

$$\left(\begin{array}{c}2n\\ n\end{array}\right)=\frac{\left(2n\right)(2n-1)(2n-2)...(n+3)(n+2)(n+1)}{n(n-1)(n-2)...3\times 2\times 1}$$$$a=\sqrt[n]{}\left(\begin{array}{c}2n\\ n\end{array}\right)$$$$\Rightarrow lna=\frac{1}{n}[ln\left(2\right)+ln(2+\frac{1}{n-1})+ln(2+\frac{2}{n-2})+...+ln(2+\frac{n-3}{3})(2+\frac{n-2}{2})(2+\frac{n-1}{1})$$$$=\frac{1}{n}\underset{i=0}{\sum ^{n-1}}ln(2+\frac{\frac{i}{n}}{1-\frac{i}{n}})$$$$\Rightarrow {lim}_{n\to \mathrm{\infty }}{a}_n={\int }_0^{1}ln(2+\frac{x}{1-x})fx$$$$={\int }_0^1[ln(2-x)-ln(1-x)]dx$$$${=(x-2)ln(2-x)-(x-1)ln(1-x))]}_0^1$$$$=2ln2=ln4\Rightarrow {lim}_{n\to \mathrm{\infty }}{a}_n=4✓$$$$$$

Answered by Frix last updated on 01/Apr/24

$$n\to \mathrm{\infty }\Rightarrow n!\to {\left(\frac{n}{\mathrm{e}}\right)}^n\sqrt{2\pi n}\Rightarrow$$$$\sqrt[n]{\left(\begin{array}{c}2n\\ n\end{array}\right)}\to \frac{4}{\sqrt[n]{\pi n}}=4$$

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