$(200!)^{.005}\approx 2(100!)^{.01}$

Question

Notice that $(200!)^{.005}\approx 2(100!)^{.01}$

Can we generalize that to $(kn)!^{\frac{1}{kn}}\approx k(n!)^{\frac{1}{n}}$?

I have tried Mathematical induction and it gets complicated.

Answer 1

Start with good old Stirling:

$n! \approx cn^{1/2}\dfrac{n^n}{e^n} $ where $c = \sqrt{2\pi}$. Then $n!^{1/n} \approx c^{1/n}n^{1/(2n)}\dfrac{n}{e} $.

Put in $kn$.

$(kn)!^{1/(kn)} \approx c^{1/(kn)}(kn)^{1/(2kn)}\dfrac{kn}{e} = c^{1/(kn)}(k)^{1/(2kn)}(n)^{1/(2kn)}\dfrac{kn}{e} $.

Therefore

$\begin{array}\\ r(k, n) &=\dfrac{(kn)!^{1/(kn)}}{n!^{1/n}}\\ &\approx \dfrac{c^{1/(kn)}(k)^{1/(2kn)}(n)^{1/(2kn)}\dfrac{kn}{e}} {c^{1/n}n^{1/(2n)}\dfrac{n}{e}}\\ & =c^{1/(kn)-1/n}(k)^{1/(2kn)}(n)^{1/(2kn)-1/(2n)}k\\ & =k\left(c^{1/(k)-1}(k)^{1/(2k)}(n)^{1/(2k)-1/(2)}\right)^{1/n}\\ & =k\left(c^{1/(k)-1}(kn)^{1/(2k)}(n)^{-1/(2)}\right)^{1/n}\\ & =k\left(c^{1/(k)-1}\right)^{1/n}(kn)^{1/(2kn)}(n)^{-1/(2n)}\\ \end{array} $

Since, for large $m$, $x^{1/m} \approx 1+\frac{\ln x}{m} $, $x^{-1/m} \approx 1-\frac{\ln x}{m} $, $m^{1/m} \approx 1+\frac{\ln m}{m} $ and $m^{-1/m} \approx 1-\frac{\ln m}{m} $,

$\begin{array}\\ r(k, n) &\approx k\left(c^{1-1/(k)}\right)^{-1/n}(kn)^{1/(2kn)}(n)^{-1/(2n)}\\ &\approx k(1-\frac{\ln(c^{1-1/(k)})}{n})(1+\frac{\ln(kn)}{2kn})(1-\frac{\ln(n)}{2n})\\ \end{array} $

so, as conjectured, $\lim_{n \to \infty} \frac1{k}r(k, n) \to 1$.

More precisely,

$\begin{array}\\ \frac1{k}r(k, n) &\approx 1+\frac{\ln(kn)}{2kn}-\frac{\ln(n)}{2n}+O(\frac1{n})\\ &\approx 1+\frac{\ln(k)}{2kn}+\frac{\ln(n)}{2kn}-\frac{\ln(n)}{2n}+O(\frac1{n})\\ &\approx 1-\frac{\ln(n)(1-1/k)}{2n}+O(\frac1{n})\\ \end{array} $

Answer 2

In the same spirit as marty cohen in his answer.

Assuming that both $lhs$ and $rhs$ are $> 1$, take logarithms to show that $${\frac{1}{kn}}\log[(kn)!]\approx \log(k)+ {\frac{1}{n}}\log(n!)$$ Using Stirling approximation for large $p$ $$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({p}\right)\right)+\frac{1}{12 p}+O\left(\frac{1}{p^3}\right)$$

$$\log(lhs)=\log (k)+\log \left({n}\right)-1+\frac{\log \left({k}\right)+\log \left({n}\right)+\log (2 \pi )}{2 k n}+O\left(\frac{1}{n^2}\right)$$

$$\log(rhs)=\log (k)+\log \left({n}\right)-1+\frac{\log (2 \pi )+\log \left({n}\right)}{2 n}+O\left(\frac{1}{n^2}\right)$$ $$\log(rhs)-\log(lhs)=\frac{(k-1) \left(\log (2 \pi )+\log \left({n}\right)\right)-\log \left({k}\right)}{2 k n}+O\left(\frac{1}{n^2}\right)$$ Using Taylor again $(x=e^{\log(x)})$ $$\frac{rhs}{lhs}=1+\frac{(k-1) \left(\log (2 \pi )+\log \left({n}\right)\right)-\log \left({k}\right)}{2 k n}+O\left(\frac{1}{n^2}\right)$$

Using your numbers $(k=2,n=100)$, the above would give $\approx 1.01437$ while the exact value is $\approx 1.01448$.