<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] Yet another way.
Lemma 1:
$$ \int_{0}^{\pi/2}\left(\cos\theta\right)^{2n}\,d\theta = \frac{\pi}{2\cdot 4^n}\binom{2n}{n} \tag{1}$$
Proof of Lemma 1: by parity and De Moivre's formula
$$ \int_{0}^{\pi/2}(\cos\theta)^{2n}\,d\theta = \frac{1}{2}\int_{-\pi/2}^{\pi/2}\left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^{2n}\,d\theta =\frac{1}{2\cdot 4^n}\sum_{k=0}^{2n}\binom{2n}{k}\int_{-\pi/2}^{\pi/2} e^{ki\theta} e^{-(2n-k)i\theta}\,d\theta $$
where the last integrals always equals zero, unless $k=n$.
Corollary:
$$ \sum_{r\geq 0}\binom{2r}{r} x^r = \frac{2}{\pi}\sum_{r\geq 0}\int_{0}^{\pi/2}(4x\cos^2\theta)^{r}\,d\theta=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{1-4x\cos^2\theta}.\tag{2} $$
Letting $\theta=\arctan u$ in the last integral we have
$$\sum_{r\geq 0}\binom{2r}{r}x^r = \frac{2}{\pi}\int_{0}^{+\infty}\frac{du}{1+u^2-4x}\stackrel{u\mapsto v\sqrt{1-4x}}{=}\frac{2}{\pi\sqrt{1-4x}}\int_{0}^{+\infty}\frac{dv}{v^2+1}=\frac{1}{\sqrt{1-4x}}.\tag{3} $$

And another one. Since by Vandermonde's identity
$$\binom{2n}{n}=\sum_{k=0}^{n}\binom{n}{k}^2 $$
the Cauchy-Schwarz inequality gives $\binom{2n}{n}\geq \frac{4^n}{n+1}$. On the other hand $\binom{2n}{n}\leq \sum_{k=0}^{n}\binom{2n}{k}=4^n$, so the radius of convergence of our power series is exactly $\frac{1}{4}$. As a consequence, the radius of convergence of
$$ f(z)=\sum_{n\geq 0}a_n z^n\quad\text{with}\; a_n=\frac{1}{4^n}\binom{2n}{n} $$
is exactly one. The coefficients $a_n$ satisfy a nice recurrence relation: since
$$ \frac{1}{4^{n+1}}\binom{2n+2}{n+1} = \frac{(2n+2)(2n+1)}{4(n+1)^2}\binom{2n}{n}=\frac{(2n+1)}{2(n+1)}\binom{2n}{n} $$
we have $2(n+1)a_{n+1}=(2n+1)a_n$, which can be turned into a differential equation. Indeed
$$ f'(z) = \sum_{n\geq 0} (n+1) a_{n+1} z^{n},\qquad z\cdot f'(z)=\sum_{n\geq 1} n a_n z^n $$
hence
$$ 2(1-z) f'(z) = f(z) $$
with $f(0)=a_0=1$. The last differential equation is a separable differential equation:
$$ \frac{f'(z)}{f(z)}=\frac{1}{2(1-z)}\quad\Longrightarrow\quad \log f(x) = \int_{0}^{x}\frac{dz}{2(1-z)} = -\frac{1}{2}\log(1-x)$$
so by exponentiating both sides of the the last identity we get
$$ \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}x^n = \frac{1}{\sqrt{1-x}} $$
as wanted.
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