I saw the following equality in an informal proof:
$\lim_{n \to \infty}\frac{\ln{n^n}}{\ln{n!!}}=\lim_{n \to \infty}\frac{\ln{n^n}-\ln(n-2)^{n-2}}{\ln{n!!}-\ln(n-2)!!}$
I did not understand it and would appreciate clarification on how it is obtained. Above is the exact notation used; I cannot help with any ambiguities about whether it is $\ln({n!!})$ or $\ln(n)!!$ although I believe it is the former. (I suppose a numerical would help figure out which is meant but I am on mobile)
Update: I would think the one who gave this proof surmised that $\lim{\ln(n-2)^{n-2}}+\ln{n^n}= \ln{n^n}$ because of the $n-2$ power, but I do not see how the expression in the denominator arises.
Some asymptotic estimates for the double factorial using Stirling.
$n!! =\prod_{j=0}^{\lfloor (n-1)/2 \rfloor} (n-2j) $.
$\begin{array}\\ (2n)!! &=\prod_{j=0}^{\lfloor (2n-1)/2 \rfloor} (2n-2j)\\ &=\prod_{j=0}^{n-1} 2(n-j)\\ &=2^n\prod_{j=0}^{n-1} (n-j)\\ &=2^nn!\\ &\approx 2^n\sqrt{2\pi n}\dfrac{n^n}{e^n}\\ &=\sqrt{2\pi n}(\dfrac{2n}{e})^n\\ \end{array} $
$\begin{array}\\ (2n+1)!! &=\prod_{j=0}^{\lfloor (2n)/2 \rfloor} (2n+1-2j)\\ &=\prod_{j=0}^{n} (2(n-j)+1)\\ &=\prod_{j=0}^{n} (2j+1)\\ &=\prod_{j=1}^{n} (2j+1)\\ &=\prod_{j=1}^{n} (2j+1)\dfrac{\prod_{j=1}^{n} (2j)}{\prod_{j=1}^{n} (2j)}\\ &=\dfrac{\prod_{j=2}^{2n+1} j}{2^n n!}\\ &=\dfrac{(2n+1)!}{2^n n!}\\ &=(2n+1)\dfrac{(2n)!}{2^n n!}\\ &\approx(2n+1)\dfrac{\sqrt{4\pi n}\dfrac{(2n)^{2n}}{e^{2n}}}{2^n \sqrt{2\pi n}\dfrac{n^n}{e^n}}\\ &=(2n+1)\dfrac{\sqrt{2}(2n)^{2n}}{2^n e^{2n}\dfrac{n^n}{e^n}}\\ &=(2n+1)\dfrac{\sqrt{2}2^nn^n}{ e^{n}}\\ \end{array} $
If $n$ is even then $n!! \approx \sqrt{\pi n}(n/e)^{n/2} $.
If $n$ is odd then, putting $(n-1)/2$ for $n$,
$\begin{array}\\ n!! &\approx n\sqrt{2}((n-1)/e)^{(n-1)/2}\\ &= n\sqrt{2}\sqrt{e/(n-1)}((n-1)/e)^{n/2}\\ &= n\sqrt{2}\sqrt{e/(n-1)}(n(1-1/n)/e)^{n/2}\\ &= \sqrt{2en^2/(n-1)}(n/e)^{n/2}(1-1/n)^{n/2}\\ &\approx \sqrt{2en}(n/e)^{n/2}e^{-1/2}\\ &\approx \sqrt{2n}(n/e)^{n/2}\\ \end{array} $