What function does $\sum_{n=0}^\infty \frac{(-1)^n(2n)!}{(1-2n)n!^24^n}\frac{x^n}{y^{2n}}$ correspond to? I have the following Taylor series, and although it looks familiar, I cannot figure out which function it corresponds to!  Does anyone recognize this Taylor series?
$$1+\frac{x}{2y^2} - \frac{x^2}{8y^4} + \frac{x^3}{16y^6}+\frac{5x^4}{128y^8}+.... = \sum_{n=0}^\infty \frac{(-1)^n(2n)!}{(1-2n)n!^24^n}\frac{x^n}{y^{2n}}$$
Thank you!

A good series to know offhand is
$$\frac{1}{\sqrt{1-4t}} = \sum_{n=0}^\infty {2n \choose n}t^n $$
which almost gets us the series we desire with $t = -\frac{x}{4y^2}$
$$\frac{1}{\sqrt{1+\frac{x}{4y^2}}} = \sum_{n=0}^\infty {2n \choose n}\left(-\frac{x}{4y^2}\right)^n $$
except for the offending term $1-2n$ in the denominator. However a small adjustment to the original series with $t = -z^{-2}$
$$\frac{1}{\sqrt{1+\frac{4}{z^2}}} = \frac{|z|}{\sqrt{z^2+4}}= \sum_{n=0}^\infty {2n \choose n}(-1)^nz^{-2n}$$
gives us the series we want via integration
$$\sqrt{z^2+4}\operatorname{sgn}(z) = \sum_{n=0}^\infty {2n \choose n}\frac{z^{1-2n}}{1-2n} \implies \frac{\sqrt{z^2+4}}{|z|} = \sum_{n=0}^\infty {2n \choose n}\frac{z^{-2n}}{1-2n}$$
by substituting $z= \frac{2y}{\sqrt{x}}$
$$\sum_{n=0}^\infty {2n \choose n}\frac{1}{1-2n}\left(-\frac{x}{4y^2}\right)^n = \boxed{\frac{\sqrt{y^2+x}}{|y|}=\sqrt{1+\frac{x}{y^2}}}$$
