TITLE: Prove that $\lim_{n \to \infty} \binom{n}{k}a^n = 0$
QUESTION [5 upvotes]: I'm working with this problem but I have no idea how to solve it. Here $k$ is fixed and $0<a<1$. 
I was trying to use that $\lim_{n \to \infty} a^n =0$ and that $\binom{n}{k}\leq\frac{n^k}{k!}$ with the $\epsilon$ definition to prove it, my intention was to show that for $N$ large enough $a^N < \frac{k!}{N^k}$ but I got nowhere. I don't know if using the definition is the best aproach.

REPLY [2 votes]: Let $t_n$ be the $n$-th term. Calculation shows that
$$\frac{t_{n+1}}{t_n}=\frac{\binom{n+1}{k}a^{n+1}}{\binom{n}{k}a^n}=\frac{n+1}{n+1-k}a=\left(1+\frac{k}{n+1-k}\right) a.$$
If $n$ is large enough, $\left(1+\frac{k}{n+1}\right)a \lt b$, for some fixed $b\lt 1$. So after a while, each time we increment $n$ by $1$, $t_n$ decreases by a factor of at least $b$.