Calculation of speed of convergence

Question

Let consider this expression $\binom{n}{k}p^k(1-p)^{n-k}$

I understand that if $n$ is very large then $(1-p)^{n-k} \to 0$ . Therefore entire expression converges to 0.

However how can I show that that the rate of convergence is $O(n^{-1/2})$?

Answer 1

Assume $0<p<1, k$ fixed.

$$\binom{n}{k}p^k(1-p)^{n-k}=\frac{n(n-1)\cdots (n-k+1)}{k!}\cdot (\frac{p}{1-p})^k\cdot (1-p)^n\sim c\cdot n^k\cdot (1-p)^n,$$ with $c=\frac{1}{k!}\cdot (\frac{p}{1-p})^k.$

Note that $n^k\cdot (1-p)^{\varepsilon n}\to 0$ for any fixed $\varepsilon>0$, so we have $$n^k\cdot (1-p)^n=n^k\cdot (1-p)^{\varepsilon n}\cdot (1-p)^{(1-\varepsilon)n}= o((1-p)^{(1-\varepsilon)n})=o(\frac{1}{x^n}),$$ with $1-p<\frac{1}{x}=(1-p)^{(1-\varepsilon)}<1$.

With $\lim_{n\to \infty}\frac{n^\alpha}{x^n}=0$ for any fixed $\alpha>0$ and $x>1$, we have $$\binom{n}{k}p^k(1-p)^{n-k}=o(\frac{1}{n^\alpha}),$$ for any fixed $\alpha>0$.