Let be $x_{n}=\int_{0}^{1}x^{n}(1-x)^{n}dx$. Find $\lim_{n \to \infty }\frac{x_{n+1}}{x_{n}}$. I proved that $\lim_{n \to \infty }x_{n}=0$ and I tried to use l'Hospital
2026-04-11 18:34:41.1775932481
Find $\lim_{n \to \infty }\frac{x_{n+1}}{x_{n}}$
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$x_n=\beta(n+1,n+1)=\frac{n!~~ n!}{(2n+1)!},~~~~~~~$ (Beta function)
So, $\lim\limits_{n\rightarrow \infty}\frac{x_{n+1}}{x_n}=\lim\limits_{n\rightarrow \infty}\frac{(n+1)!~~(n+1)!}{(2n+3)!}\times \frac{(2n+1)!}{n!~~ n!}=\lim\limits_{n\rightarrow \infty}\frac{(n+1)~~(n+1)}{(2n+3)~~(2n+2)}=\frac{1}{4}$.