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 $n \geq k$?



Asymptotic behaviour of binomial term
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 $n=k/p$.


