Asymptotic behaviour of binomial term Is it true that that ${{n}\choose{k}} p^k (1-p)^{n-k}$ is dominated by $\frac{1}{n}$, at least for $k$ sufficiently big?
EDIT: I saw that the question was absolutely not stated as I intended. The real question is: given a real number $p \in (0, 1)$, assuming $k$ fixed big enough, is there some constant $C > 0$ independent from $k$ such that $n \cdot {{n}\choose{k}} q^k (1-q)^{n-k} \leq C$ for every $n \geq k$?
 A: Assuming $0 \le p \le 1$, $n$ and $k$ nonnegative integers with $n \ge k$, 
${n \choose k} p^k(1-p)^{n-k} = \mathbb P(X = k)$ where $X$ is a binomial random variable with parameters $n$ and $p$ (i.e. the number of successes in $n$ independent Bernoulli trials with probability $p$ of success in each).
I think you want to take $n \to \infty$ with fixed $k$.  Of course, if $p = 0$ or $p=1$ the probability is $0$ for $1 \le k \le n-1$, so let's suppose $0 < p < 1$.  Now ${n \choose k}$ is a polynomial in $n$, while $(1-p)^{n-k}$ decays exponentially, and therefore faster than any negative power of $n$.  Thus it is true that $\mathbb P(X=k) < 1/n$ for sufficiently large $n$.
A: Say $k=np$ (imagine $p=1/2$), then that is about $C/\sqrt{n}$ for some constant $C$.  This is by (for example) a normal approximation to the underlying binomial random variable, or (perhaps more to the point) recognizing that the binomial random variable has standard deviation about $C\sqrt{n}.$
This ($1/\sqrt{n}$) is the biggest the sum gets.  If $|k-np|/\sqrt{n}$ is much larger than 1, then this expression really really quickly falls to $0$.  (See many different bounds for this, but perhaps just bound it by the normal approximation.)
You may find useful the bounds $(n/k)^k \leq {n \choose k} \leq (en/k)^k,$ which almost always help clear things like this up considerably.
A: Robert's answer is correct, if you want an asymptotic answer. However, for the question as stated in the edit (that is for all $n\ge k$...) the answer is no.
Take for example the case of $p=\frac12$. For any $k$, if you take $n=2k$ then ${n \choose k}p^k(1-p)^{n-k}=\frac{(2k)!}{k!k!2^{2k}}$ which is roughly $\frac{1}{\sqrt{\pi k}}$ by Stirling's formula. This is, of course, bigger then $\frac{1}{n}$.
The same also works for other values of $0<p<1$ by taking $n=k/p$.
