Prove or disprove that $\lim_{n \to \infty} {n \choose k} / a^n = 0$ for all $k \in \mathbb{Z_+}, a \in \mathbb{R}: a> 1$.

Question

Question: Prove or disprove that $\lim_{n \to \infty} {n \choose k} / a^n = 0$ for all $k \in \mathbb{Z_+}, a \in \mathbb{R}: a> 1$.

Comment: When I was falling asleep last night I was thinking about increasing integer sequences like triangle numbers $(0,1,3,6,10,15,\cdots)$ and tetrahedral numbers $(0,1,4,10,20,35,\cdots)$ or in short any integer sequences of the type $\left({0 \choose k}, {1 \choose k}, {2 \choose k}, \cdots\right)$ for some integer $k$ and I was wondering if for some $k$ that there is exponential growth at the limit, hence my question. I suppose not, but I could not prove it directly.

What if the $k$ constant increases as a function of $n$, for example if $k = \lceil n/2 \rceil$ (giving the sequence $\left({0 \choose 0}, {1 \choose 1}, {2 \choose 1}, {3 \choose 2}, \cdots\right)$)?

If the constant increases as suggested, the growth rate is roughly like $4^n$, so the answer will depend on $b$, — André Nicolas, May 09 '14 at 20:38

score 2 · Accepted Answer · answered May 09 '14 at 20:39

2

Note that for $n\geq a$ we have $\binom{n}{a}\leq \frac{n^a}{a!}.$ So $\binom{n}{a}$ growths, at most, polynomically while $b^n$ growths exponentially. Thus, $\displaystyle \lim_{n\rightarrow \infty} \frac{\binom{n}{a}}{b^n}=0.$

answered May 09 '14 at 20:39

mfl

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Prove or disprove that $\lim_{n \to \infty} {n \choose k} / a^n = 0$ for all $k \in \mathbb{Z_+}, a \in \mathbb{R}: a> 1$.

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