Verifying asymptotic of sequence

Question

Let $p\in (0,1), q := 1-p$. Let $k_n$ be a sequence such that $k_n - np \sim c\sqrt{npq}$. I want to show that $$\sqrt{\frac {n}{k_n (n-k_n)}} \sim \frac 1 {\sqrt{npq}}$$ as $n\to\infty$.

Answer 1

First note that $\frac {k_n} n \to p$. Then write ${\sqrt {npq} \frac {\sqrt n} {\sqrt { {k_n(n-k_n)}}}}$ as $\sqrt {pq} \frac 1 {a_n}$ where $a_n=\sqrt { \frac {k_n} n (1-\frac {k_n}n)}$. Can you now see that ${\sqrt {npq} \frac {\sqrt n} {\sqrt { {k_n(n-k_n)}}}} \to 1$?

[$\frac {\sqrt {pq}} {\sqrt {p(1-p)}}=1$ because $q=1-p$].

Answer 2

Consider $$A=\sqrt{\frac {n}{k_n (n-k_n)}}\implies A^2 =\frac {n}{k_n (n-k_n)}=\frac n { (c \sqrt{n p q}+n p)(n-np-c \sqrt{n p q })}$$ Now, let $n=\frac 1 {x^2}$ to make $$A^2=-\frac{x^2}{\left(c x \sqrt{p q}+p-1\right) \left(c x \sqrt{p q}+p\right)}$$ Expand the denominator to get $${\left(c x \sqrt{p q}+p-1\right) \left(c x \sqrt{p q}+p\right)}=\left(p^2-p\right)+x \left(2 c p \sqrt{p q}-c \sqrt{p q}\right)+c^2 p q x^2$$ Now, long division to make $$A^2=\frac{x^2}{p-p^2}+\frac{c (2 p-1) \sqrt{p q}}{(p-1)^2 p^2}x^3+O\left(x^4\right)$$ Factor the $x^2$ and continue with Taylor series to get $$A=\frac x{\sqrt{{p-p^2}}} +\frac{c (1-2 p) q }{2 (p-1) \sqrt{p q(p-p^2)}}x^2+O\left(x^3\right)$$ Using $q=1-p$ and truncating, then replace $x$ by $\frac 1 {\sqrt n}$ to get your result.