A continued fraction for the reciprocal of Gauss's constant

This preprint from a couple days ago doesn't seem entirely rigorous (and gives the wrong value for Gauss's constant) but gave me enough guidance for a pretty clean proof. The key is using Pincherle’s theorem (see Theorem $2$ on pg $5$).

Define $$I_n=\int_0^2x^{n+1/4}(2-x)^{1/2}(4-x)^{1/4}dx $$ Because $x^{n-3/4}(2-x)^{3/2}(4-x)^{5/4}$ evaluates to $0$ at the boundary points and its derivative is

$$\Big[ (n+2)x^n - (6n+4)x^{n-1} + 2(4n-3)x^{n-2} \Big] x^{1/4}(2-x)^{1/2}(4-x)^{1/4}$$ It follows that $$(n+2)I_n=(6n+4)I_{n-1}-2(4n-3)I_{n-2}$$ Now define $$u_{2n}=I_n(n+2)! \qquad u_{2n-1}=u_{2n}-I_{n-1}(4n+1)(n+1)! $$ Using our $I_n$ recurrence along with some basic algebra and rearranging, we have $$u_{2n}=u_{2n-1}+(4n+1)u_{2n-2}$$ $$u_{2n-1}=u_{2n-2}+(2n+2)u_{2n-3}$$ The only remaining point is to show that $u_n$ is the minimal solution which is necessary for Pincherle's. Define $$J_n=\int_2^4x^{n+1/4}(x-2)^{1/2}(4-x)^{1/4}dx$$ and construct $v_n$ analogously from $J_n$ just as $u_n$ was constructed from $I_n$. Indeed, $v_n$ satisfies the same recurrence. However, since $I_n^{1/n}\to 2$ and $J_n^{1/n}\to 4$, we have $u_{2n}/v_{2n}=I_n/J_n\to0$ so $u_n$ is minimal. Thus by Pincherle's theorem, the continued fraction converges to $$-\frac{u_0}{u_{-1}}=\frac{2I_0}{I_{-1}-2I_0}$$ And it simply remains to evaluate two elementary beta integrals $$I_0=\int_0^2x^{1/4}(2-x)^{1/2}(4-x)^{1/4}dx=\frac{\pi}{\sqrt2}$$ $$I_{-1}=\int_0^2x^{-3/4}(2-x)^{1/2}(4-x)^{1/4}dx=\sqrt{2}\pi + \frac{\pi^{3/2}}{\Gamma\left(\frac{3}{4}\right)^2}$$