Calculate limit involving binomial coefficient

How can I calculate this limit. Let $p\in[0,1]$ and $k\in\mathbb{N}$.

$$\lim\limits_{n\to\infty}\binom{n}{k}p^k(1-p)^{n-k}.$$

Any idea how to do it ?

A sketch of proof: The element is $$\frac{n(n-1)(n-2)\dots(n-k+1)}{k!} \left(\frac p{1-p}\right)^k (1-p)^n \\= A_k n(n-1)(n-2)\dots(n-k+1)(1-p)^n$$ for a certain $A_k$ which does not influence the convergence.
When $n\to\infty$ this is equivalent to $$a_n = A_k n^k(1-p)^n$$
But $a_{n+1}/a_n = \left(\frac{n+1}{n}\right)^k(1-p) < 1-\frac p2$ as soon as $p\neq 1$, for $n$ big enough. So the limit is 0 in that case (prove it using induction).
• It's not quite enough to say that $a_{n+1}/a_n < 1$; that is true for any decreasing sequence, whether it converges to zero or not. We get zero here because the polynomial growth of $n^k$ is defeated by the exponential decay of $(1-p)^n$. – user21467 Oct 21 '14 at 23:53
Let $X_n$ be a random variable with binomial distribution $B(n,p)$. (That is, it's the number of successes in $n$ independent attempts, where $p$ is the probability of success.) Then $$\binom nk p^k (1-p)^{n-k} = P(X_n=k) \le P(|X_n-np|\ge np-k) \stackrel{(\ast)}{\le} \frac{\operatorname{Var} X_n}{(np-k)^2} = \frac{np(1-p)}{(np-k)^2}$$ which tends to zero as $n\to\infty$ because the top has order $n$ and the bottom has order $n^2$. (The step ($\ast$) is Chebyshev's inequality.)