<s>[INST] 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?
 [/INST] Use the Duplication formula for Gamma function
$$
\Gamma \left( {2\,z + 1} \right) = \Gamma \left( {2\,\left( {z + 1/2} \right)} \right)
 = 2^{\,2\,z} \frac{{\Gamma \left( {z + 1} \right)\Gamma \left( {z + 1/2} \right)}}{{\Gamma \left( {1/2} \right)}}
$$
So
$$
\begin{array}{l}
 \left( \begin{array}{c}
 2r \\  r \\ 
 \end{array} \right) = \frac{{2r!}}{{r!r!}} = \frac{{\Gamma \left( {2r + 1} \right)}}{{\Gamma \left( {r + 1} \right)\Gamma \left( {r + 1} \right)}} =  \\ 
  = 2^{\,2\,r} \frac{{\Gamma \left( {r + 1} \right)\Gamma \left( {r + 1/2} \right)}}{{\Gamma \left( {r + 1} \right)\Gamma \left( {r + 1} \right)}}
 = 4^{\,\,r} \frac{{\Gamma \left( {r + 1/2} \right)}}{{\Gamma \left( {r + 1} \right)\Gamma \left( {1/2} \right)}} =  \\ 
  = 4^{\,\,r} \frac{{\left( {r - 1/2} \right)!}}{{r!\left( { - 1/2} \right)!}} = 4^{\,\,r} \left( \begin{array}{c}
 r - 1/2 \\ 
 r \\ 
 \end{array} \right) =  \\ 
 \left| {\;0 \le r \in \Bbb{Z}} \right. \\ 
  = \left( { - 1} \right)^{\,\,r} 4^{\,\,r} \left( \begin{array}{c}
  - 1/2 \\  r \\ 
 \end{array} \right) \\ 
 \end{array}
$$
</s>