Asymptotic behaviour of Gamma functions I would like to simplify or get a simpler approximation of the following ratio involving the  Gamma function
\begin{equation}\tilde{\gamma}(K)\triangleq \frac{1}{\sqrt{K}}\frac{\Gamma\left(\frac{3K}{2}+\frac{1}{2}\right)}{\Gamma\left(\frac{3K}{2}\right)} \qquad K\in\mathbb{N} \end{equation}
I have to do it because of two reasons:

*

*I have to evaluate $\tilde{\gamma}$ for large values of $K$. Currently I can push my PC up to $K=100$ by evaluating
\begin{equation*}\tilde{\gamma}(K)=\exp\left\{-\frac{\log(K)}{2}+\log\left[\Gamma\left(\frac{3K}{2}+\frac{1}{2}\right)\right]-\log\left[\Gamma\left(\frac{3K}{2}\right)\right]\right\}\tag{1}\end{equation*}
but this is not satisfying since in practice my $K$s are in the order of $10^4$;

*I want to know if it exists, and in such case get its value, a precise limit as $K\to \infty$. The graph of $\tilde{\gamma}$ for $K=1,\dots,100$ suggests that probably there is a specific limit value around $1$ (in particular, it seems $\sqrt{3/2}$). In that case, it would be nice to know the limit value because it will allow me to replace the computation of $\tilde{\gamma}$ with just a fixed value when $K$ is "large" (in a sense that should be clarified).

I know that there are plenty of useful properties that can be applied, but unfortunately I end up with a quite complex result. The main problem is that my "semplifications" dont clarify if $\tilde{\gamma}$ has a limit. Down below you can find my strategy to rearrange $\tilde{\gamma}$ in a more suitable form.
Even simplification
Let $3K/2=\bar{K}$ for some integer $\bar{K}$. This happens every time $K$ is even, and so the name "even simplification".
In this special case holds the following simplification
\begin{equation*}\Gamma\left(\frac{3K}{2}+\frac{1}{2}\right)=\Gamma\left(\bar{K}+\frac{1}{2}\right)=\frac{(2\bar{K}-1)!!}{2^\bar{K}}\sqrt{\pi}\end{equation*}
and also the simplification
\begin{equation*}\Gamma\left(\frac{3K}{2}\right)=\Gamma\left(\bar{K}\right)=(\bar{K}-1)!\end{equation*}
so that
\begin{equation*}\tilde{\gamma}(K)=\frac{1}{\sqrt{K}}\frac{(2\bar{K}-1)!!}{(\bar{K}-1)! \,2^\bar{K}}\sqrt{\pi}\end{equation*}
Now, in order to avoid the factorial and the double factorial, I use the following products
\begin{equation*}
\begin{aligned}
(2\bar{K}-1)!!&=\prod_{j=1}^{\bar{K}} (2j-1)\\
(\bar{K}-1)!&=\frac{1}{\bar{K}}\prod_{j=1}^{\bar{K}} j\\
2^\bar{K} &= \prod_{j=1}^{\bar{K}} 2
\end{aligned}
\end{equation*}
which allows to write the incriminated ratio in the following very nice form
\begin{equation*}
\tilde{\gamma}(K)=\frac{1}{\sqrt{K}}\left(\bar{K}\prod_{j=1}^\bar{K} \frac{2j-1}{2j}\right)_{\bar{K}=3K/2} \sqrt{\pi}
\end{equation*}
as a final step, in order to make the expression overflow-resistant, I write the result in its exp/log version
\begin{equation*}
\tilde{\gamma}(K)=\exp\left(-\frac{\log(K)}{2}+\log(\bar{K})+\sum_{j=1}^\bar{K} \log(2j-1)-\log(2j)\right)_{\bar{K}=3K/2} \sqrt{\pi}
\end{equation*}
Odd simplification
With some minor adjustments to the derivation above I get the following results for the remaining case $K$ odd
\begin{equation*}
\tilde{\gamma}(K)=\frac{1}{\sqrt{K}}\left(\prod_{j=1}^\bar{K} \frac{2j}{2j-1}\right)_{\bar{K}=(3K+1)/2} \frac{1}{\sqrt{\pi}}
\end{equation*}
and so
\begin{equation*}
\tilde{\gamma}(K)=\exp\left(-\frac{\log(K)}{2}-\sum_{j=1}^\bar{K} \log(2j-1)-\log(2j)\right)_{\bar{K}=(3K+1)/2} \frac{1}{\sqrt{\pi}}
\end{equation*}
Question
My simplifications seem to agree pretty well with $(1)$ (with the advantage that they works also above $K=100$), but I'm still not be able to say if there is a limit value for $\tilde{\gamma}$, so my questions are the following:

*

*does the limit exist for $\tilde{\gamma}$?

*if the limit exist, what is its value?

 A: I've found a trivial proof of the fact
\begin{equation*}\lim_{K\to\infty} \tilde{\gamma}=\sqrt{\frac{3}{2}}\end{equation*}
by using the Stirling's formula applied to $\Gamma(3/2\,K+1/2)$ and $\Gamma(3/2 K)$.
A: You can use the standard asymptotic expansion for the ratio of two gamma functions to deduce
$$
\widetilde{\gamma}(K) \sim \sqrt {\frac{3}{2}} \left( {1 - \frac{1}{{12K}} + \frac{1}{{288K^2 }} + \frac{5}{{{\rm 3456}K^3 }} -  \ldots } \right),
$$
as $K\to +\infty$. The corresponding factorial series is
$$
\tilde \gamma (K) \!=\! \sqrt {\frac{3}{2}}\! \left( {1 \!-\! \frac{1}{{12(K + 1)}} \!-\! \frac{{23}}{{288(K + 1)(K + 2)}} \!-\! \frac{{535}}{{{\rm 3456}(K + 1)(K + 2)(K + 3)}} \!-\!  \ldots } \right)
$$
which converges for $K>0$.
A: Since @Gary already gave the solution, if your problem is numerical (as you tagged), you can have a very good approximation using the corresponding $[n,n+1]$ Padé approximant $P_n$ of the logarithm.
For example
$$\log\left({\gamma}\right)=\frac{1}{2} \log \left(\frac{3}{2}\right)-\frac 1{K^2}\frac {\frac{1}{12 }+\frac{4121}{140616
   K^2} } {1+\frac{241}{651 K^2}+\frac{257}{58590 K^4} }$$ whose error is $\sim \frac 1 {50000\,K^9}$.
This is good even for small values of $K$. Taking the exponential, the relative error is less than $10^{-5}$% as soon as $K \geq 2$.
Another advantage of this type of approximation is that, if you need to inverse it, using a simpler one, you just need to solve for $K$
$$\frac{9 K}{1+54 K^2}=\log \left(\frac{3}{2 \gamma ^2}\right)$$
For example, if $\gamma=1.215$, the above gives, as a real, $K=10.4299164$ while the exact solution is   $K=10.4299182$
