There is a $p\in(0,1)$ such that $\lim_{n\to\infty}\binom{2n}n p^n\in (0,\infty)$?

Question

There is a $p\in (0,1)$ such that $\lim_{n\to\infty}\binom{2n}n p^n\in (0,\infty)$?

I was playing in Wolfram Mathematica with the above limit for $p\in(0,1)$. Surprisingly I find that when $p\in(0,1/4)$ the above limit is zero, and when $p\in(1/3,1)$ the limit is infinity.

Then I tried to find a value of $p$ such that the limit would be finite and different of zero. To do this I tried to write the limit in terms of the gamma function, then I find the expression

$$\binom{2n}np^n=\frac{(4p)^n\Gamma(n+1/2)}{\sqrt\pi(n+1)!}$$

but this doesnt simplify my search. An idea is to find some real $\alpha,\beta$ such that

$$\alpha \beta^n\sim_{\infty} \frac{\Gamma(n+1/2)}{(n+1)!}$$

but I dont know what to do from here to know if such $p$ values exists and, if they exists, how to find them. Can some one put some light in this question? Thank you in advance.

Answer 1

This is related to the Catalan numbers: you have

$$ \binom{2n}{n} = (n+1) C_n $$

According to Wikipedia, it's known that

$$ C_n \sim \frac{4^n}{n^{3/2} \sqrt{\pi}} $$

(in the sense that the limit of their ratio converges to $1$)

In particular, this means

$$\lim_{n \to \infty} \binom{2n}{n} \frac{\sqrt{n}}{4^n} = \frac{1}{\sqrt{\pi}}$$

and thus no such $p$ can exist.