Evaluating $ \sum_{r=0}^\infty { 2r \choose r} x^r $ I recently came across this question during a test

Evaluate the following $$ \sum_{r=0}^ \infty \frac {1\cdot 3\cdot 5\cdots(2r-1)}{r!}x^r $$

I converted it to the follwing
into $\sum_{r=0}^\infty  { 2r \choose r} x^r $
Then I  tried using Beta function to further expand the binomial coefficient and tried to write it as sum of integrals but a pesky "r" term ends up being multiplied
How can we evaluate this with and without using calculus?

The following approach is somewhat like taking a sledgehammer to crack a nut, but it can also conveniently used in more challenging situations. We use the coefficient of operator $[x^r]$ to denote the coefficient of $x^r$ of a series. This way we can write
\begin{align*}
\binom{2r}{r}&=[x^{r}](1+x)^{2r}\tag{1}\\
\end{align*}
We now use a clever change of variables stated as rule 5 in section 1.2 of Integral Representation and the Computation of Combinatorial Sums by G. P. Egorychev. Rule 5 adapted for this special case is:
\begin{align*}
[x^r]f^{r}(x)
&=[y^r]\left.\frac{f(x)}{f(x)-xf^{\prime}(x)}\right|_{x=g(y)}\tag{2}
\end{align*}
Here we have $f(x)=(1+x)^2$ and $g=g(y)$ is the inverse function of
\begin{align*}
\frac{x}{f(x)}=\frac{x}{(1+x)^2}=y
\end{align*}
We obtain
\begin{align*}
&yx^2+(2y-1)x+y=0\\
&x=\frac{1}{2y}\left(1-2y\pm\sqrt{1-4y}\right)\tag{3}\\
\end{align*}
We take from (3) the root $x$ with the minus sign, since this one represents a power series.

We obtain from (1),(2) and (3)
\begin{align*}
\color{blue}{\binom{2r}{r}}&=[x^{r}](1+x)^{2r}\\
&=[y^r]\left.\frac{(1+x)^2}{(1+x)^2-x\cdot 2(1+x)}\right|_{x=\frac{1}{2y}\left(1-2y-\sqrt{1-4y}\right)}\\
&=[y^r]\left.\frac{1+x}{1-x}\right|_{x=\frac{1}{2y}\left(1-2y-\sqrt{1-4y}\right)}\\
&\,\,\color{blue}{=[y^r]\frac{1}{\sqrt{1-4y}}}\tag{4}\\
\end{align*}

Since we can write a power series $A(x)=\sum_{r=0}^{\infty}a_r x^r$ as
\begin{align*}
A(x)=\sum_{r=0}^{\infty}a_r x^r=\sum_{r=0}^{\infty}\left([y^r]A(y)\right) x^r
\end{align*}
we obtain from (4)

\begin{align*}
\color{blue}{\sum_{r=0}^\infty { 2r \choose r} x^r} =\sum_{r=0}^\infty \left([y^r]\frac{1}{\sqrt{1-4y}}\right)x^r=\color{blue}{\frac{1}{\sqrt{1-4x}}}
\end{align*}

