Equivalent of a recurrence sequence $I_n=\frac{4n-3}{4n}I_{n-1}$ I have this sequence $(I_n)$ defined by $I_0=\frac{\pi}{2\sqrt2}$ and for all $n\ge1$, $I_n=\frac{4n-3}{4n}I_{n-1}$ (for those interested, it comes from $I_n=\int_0^1 \frac{t^n}{t^{3/4}(1-t)^{1/4}}{\rm d}t$).
I know how to find the order of decay of the sequence :
$$\ln\left(\frac{I_n}{I_{n-1}}\right) = -\frac{3}{4n}+O(1/n^2)$$
so, by summation and exponentiation,
$$I_n\sim \frac{K}{n^{3/4}}$$
What I'd like to know is : is there a way to compute the constant $K$ in a simple way (i.e. no Gamma function, just elementary analysis) ? It looks like the Wallis sequence, but it's more difficult to insert the missing termes in the factorials...
Thanks for reading.
\bye
 A: By Euler's Beta function
$$ I_n = \int_{0}^{1} t^{n-3/4} (1-t)^{-1/4}\,dt = \frac{\Gamma\left(n+\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)}{\Gamma(n+1)} $$
hence
$$ \color{red}{K}=\lim_{n\to +\infty} n^{3/4} I_n = \Gamma\left(\frac{3}{4}\right)\lim_{n\to +\infty}\frac{\Gamma\left(n+\frac{1}{4}\right)n^{3/4}}{\Gamma(n+1)}=\color{red}{\Gamma\left(\frac{3}{4}\right)}=\left(\frac{\pi}{2}\right)^{1/4}\sqrt{\text{AGM}(1,\sqrt{2})}. $$
A: If you use Pochhammer symbols and $I_0=\frac{\pi}{2\sqrt2}$
$$I_n=\frac{4n-3}{4n}I_{n-1}\quad \implies I_n=\frac{\pi }{8 \sqrt{2}}\frac{\left(\frac{5}{4}\right)_{n-1}}{(2)_{n-1}}=\frac{\pi }{8 \sqrt{2}\,\Gamma \left(\frac{5}{4}\right)}\frac{\Gamma \left(n+\frac{1}{4}\right)}{\Gamma (n+1)}$$ Now, for large values of $n$, using Stirling approximation
$$\frac{\Gamma \left(n+\frac{1}{4}\right)}{\Gamma (n+1)}=\frac 1 {n^{3/4}}\sum_{k=0}^\infty \frac {a_k}{n^k}$$ the first $a_k$ making the sequence
$$\left\{1,-\frac{3}{32},-\frac{7}{2048},\frac{231}{65536},\frac{7931}{8388608},-\frac
   {250173}{268435456},-\frac{9330387}{17179869184},\cdots\right\}$$ Using the above truncated series, the relative error is smaller than $10^{-6}$ as soon as $n>2$.
If you need something shorter, transforming the series into a simple Padé approximant would give
$$I_n=\frac{\pi }{8 \sqrt{2}\,\Gamma \left(\frac{5}{4}\right)} \frac 1 {n^{3/4}}\Bigg[\frac{192n-25 } {192n-7 }+O\left(\frac{1}{n^3}\right)\Bigg]$$
