Reference request for a series that involves particular values of the gamma function This week I tried to combine different integral representations for the Apéry's constant, to compute other different. For a computation that I've evoked I need $(1)$, if possible, in terms of well-known functions. The series in $(1)$ involve particular values of the gamma funtion (denoted as) $\Gamma(s)$

Question. Is it known the closed-form of $$\sum_{k=0}^\infty(-1)^k\frac{\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k}{2}+1\right)}y^{k-1}\tag{1}$$
  in terms of known functions, for $0<y<1$? If it is in the literature please refer it, and I try to search and read it from the literature. Many thanks.

I believe that a CAS could to compute in closed-form the partial sums of $(1)$.
 A: Here's my derivation.
As is my usual practice,
when a $(-1)^k$
occurs in a sum,
I split it into
even and odd parts.
I Use
$\Gamma(k+\frac12)
=\dfrac{\sqrt{\pi}(2k)!}{4^kk!}
$,
so
$\Gamma(k+\frac32)
=\dfrac{\sqrt{\pi}(2k+2)!}{4^{k+1}(k+1)!}
$,
and
(these next to
according to Wolfy)
$\sum_{k=0}^{\infty} \binom{2k}{k}x^k
=\dfrac1{\sqrt{1-4x}}
$
and
$\sum_{k=0}^{\infty} \dfrac{k! (k + 1)! x^k}{(2 k + 2)!}
 = \dfrac{2 \sin^{-1}(\frac{\sqrt{x}}{2})}{\sqrt{4-x}}
 $
$\begin{array}\\
f(y)
&=\sum_{k=0}^\infty(-1)^k\dfrac{\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k}{2}+1\right)}y^{k-1}\\
&=\sum_{k=0}^\infty\left(\dfrac{\Gamma\left(\frac{2k+1}{2}\right)}{\Gamma\left(\frac{2k}{2}+1\right)}y^{2k-1}-\dfrac{\Gamma\left(\frac{2k+1+1}{2}\right)}{\Gamma\left(\frac{2k+1}{2}+1\right)}y^{2k+1-1}\right)\\
&=\sum_{k=0}^\infty\left(\dfrac{\Gamma\left(k+\frac12\right)}{\Gamma\left(k+1\right)}y^{2k-1}-\dfrac{\Gamma\left(k+1\right)}{\Gamma\left(k+\frac32\right)}y^{2k}\right)\\
&=\sum_{k=0}^\infty\left(\frac{\dfrac{\sqrt{\pi}(2k)!}{4^kk!}}{k!}y^{2k-1}-\frac{k!}{\dfrac{\sqrt{\pi}(2k+2)!}{4^{k+1}(k+1)!}}y^{2k}\right)\\
&=\sum_{k=0}^\infty\left(\dfrac{\sqrt{\pi}(2k)!}{4^kk!^2}y^{2k-1}-\dfrac{4^{k+1}k!(k+1)!}{\sqrt{\pi}(2k+2)!}y^{2k}\right)\\
&=\dfrac{\sqrt{\pi}}{y}\sum_{k=0}^\infty\dfrac{(2k)!}{4^kk!^2}y^{2k}-\dfrac{4}{\sqrt{\pi}}\sum_{k=0}^\infty\dfrac{4^{k}k!(k+1)!}{(2k+2)!}y^{2k}\\
&=\dfrac{\sqrt{\pi}}{y}\sum_{k=0}^\infty\dfrac{(2k)!}{k!^2}(y^2/4)^{k}-\dfrac{4}{\sqrt{\pi}}\sum_{k=0}^\infty\dfrac{k!(k+1)!}{(2k+2)!}(4y^2)^{2k}\\
&=\dfrac{\sqrt{\pi}}{y}\dfrac1{\sqrt{1-4(y^2/4)}}-\dfrac{4}{\sqrt{\pi}} \dfrac{2 \sin^{-1}(\frac{\sqrt{4y^2}}{2})}{\sqrt{4-4y^2}}\\
&=\dfrac1{\sqrt{1-y^2}}\left(\dfrac{\sqrt{\pi}}{y}-\dfrac{4\sin^{-1}(y)}{\sqrt{\pi}}  \right)\\
\end{array}
$
A: Wolfram Mathematica can sum it for you, with the result
$$
\frac{\pi -2 \sin ^{-1}(y)}{\sqrt{\pi } y \sqrt{1-y^2}}
$$
