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Let $C_n$ denote the $n$th Catalan number. Prove that $\lim\limits_{n\to\infty} \dfrac{C_n}{4^n} = 0.$

I know that polynomials generally grow more slowly than exponential functions, so if $P(n)$ is a polynomial in $n$ of finite degree, then $\lim\limits_{n\to \infty} \dfrac{P(n)}{4^n} = 0.$ I also know that $C_n = \dfrac{1}{n+1} {2n\choose n} = \cdot \dfrac{(2n)(2n-1)\cdots (n+1)}{(n+1)! },$ which can be simplified further by using the fact that for even $n$, $2^{n/2+1}(n+1)n(n-1)\cdots (n/2) = (n+1)(2n)(2n-2)\cdots n$ and a similar expression can be derived for odd $n$. However, I'm not sure how to continue from here. Clearly, I cannot get a product of finitely many $n$'s; the number of $n$'s multiplied varies arbitrarily.

Bernard
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Fred Jefferson
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3 Answers3

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We have $\displaystyle 4^n = (1+1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} \ge \binom{2n}{n} = (n+1)C_n$. Then $\displaystyle \frac{C_n}{4^n} \le \frac{1}{n+1}$.

jjagmath
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We know from Stirling's formula that $$n!\sim\sqrt{2\pi n}\left(\frac ne\right)^n.$$ Then $\binom{2n}n=\frac{(2n)!}{(n!)^2}$ means $$C_n\sim\frac1{n+1}\frac{4^n}{\sqrt{\pi n}}.$$ Then $\lim_{n\to\infty}\frac{C^n}{4^n}=\lim_{n\to\infty}\frac1{(n+1)\sqrt{\pi n}}=0.$

dua
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Inequality $(10)$ from this answer says $$ \frac{4^n}{\sqrt{\pi\!\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}} $$ which says that $$ \frac{C_n}{4^n}\le\frac1{(n+1)\,\sqrt{\pi\!\left(n+\frac14\right)}} $$

robjohn
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