Show $\dfrac{(n!)^{a}n^{bn}}{\left( (2n)!\right)^{c}}\sim Kn^{(a+b-2c)n}n^{\frac{a-c}{2}}\left(e^{2c-a}2^{-2c} \right)^{n}$

Question

I would like to show: $$\dfrac{(n!)^{a}n^{bn}}{\left( (2n)!\right)^{c}}\sim Kn^{(a+b-2c)n}n^{\frac{a-c}{2}}\left(e^{2c-a}2^{-2c} \right)^{n} \mbox{ with } k=2^{\frac{a}{2}-c}\pi^{\frac{a-c}{2}}$$

My proof:

note that :

$$n!\sim \left( \dfrac{n}{e}\right)^{n}\sqrt{2\pi n}$$

\begin{align*} \dfrac{(n!)^{a}n^{bn}}{\left( (2n)!\right)^{c}}&\sim \dfrac{\left(\left( \dfrac{n}{e}\right)^{n}\sqrt{2\pi n}\right)^{a}n^{bn}}{ \left(\left( \dfrac{2n}{e}\right)^{2n}\sqrt{2\pi 2n}\right)^{c}} \\ &\sim \dfrac{n^{a+b-2c}}{(4e)^{n}}n^{n\left(\dfrac{a}{2}-c \right)}(2\pi)^{n\left(\frac{a}{2}-c \right)}\dfrac{1}{(4e)^{n}} \\ \end{align*}

• I can't get what they annonced

Answer 1

$$ \dfrac{(n!)^{a}n^{bn}}{\left( (2n)!\right)^{c}}\sim \dfrac{\left(\left( \dfrac{n}{e}\right)^{n}\sqrt{2\pi n}\right)^{a}n^{bn}}{ \left(\left( \dfrac{2n}{e}\right)^{2n}\sqrt{2\pi 2n}\right)^{c}} $$ so we have $$ \frac{\left(\frac{n}{e}\right)^{an}}{\left(\frac{n}{e}\right)^{2nc}}\sqrt{\frac{(2\pi n)^a}{(2\pi 2n)^c}}n^{bn}=n^{(a-2c)n}\mathrm{e}^{(2c-a)n}2^{-2nc}\sqrt{(2\pi)^{a-c}}\sqrt{n^{a-c}}\sqrt{2^{-c}}n^{bn} $$ this leads to $$ \sqrt{(2\pi)^{a-c}}\sqrt{2^{-c}}=\pi^{\frac{a-c}{2}}2^{\frac{a}{2}-c}\\ n^{(a-2c)n}n^{bn} = n^{(b+a-2c)n}\\ \mathrm{e}^{(2c-a)n}2^{-2nc} = \left(\mathrm{e}^{2c-a}2^{-2c}\right)^n $$

Give more steps of what you have tried and compare with above.