What is $\sum\limits_{n=0}^\infty \binom{2n}{n}\frac{1}{x^n}$ for $x > 4$. 
What is $\sum\limits_{n=0}^\infty \binom{2n}{n}\frac{1}{x^n}$ for $x > 4$.

Here is what I got so far (using Cauchy's integral formula) :
$$\sum\limits_{n=0}^\infty \binom{2n}{n}\frac{1}{x^n} ~=~ \sum\limits_{n=0}^\infty \left(\frac{1}{2\pi i}\int_\mathbb{T} \frac{(1+z)^{2n}}{z^{n+1}}\; dz\right) \cdot \frac{1}{x^n} \\=~ \frac{1}{2\pi i}\sum\limits_{n=0}^\infty   \frac{1}{x^n}\int_{\mathbb{T}} \frac{1}{z}\cdot \left( \frac{(1+z)^2}{z}\right)^n \; dz \\=~ \frac{1}{2\pi i}  \int_{\mathbb{T}} \frac{1}{z} \sum\limits_{n=0}^\infty \left( \frac{(1+z)^2}{xz}\right)^n \; dz \\=~ \frac{1}{2\pi i}\int_\mathbb{T} \frac{1}{z} \cdot \left( \frac{1}{1- \left(\frac{(1+z)^2}{xz}\right)}\right) \; dz  \\=~ \frac{1}{2\pi i}\int_\mathbb{T} \frac{1}{z} \cdot \left(\frac{xz}{xz- (1+z)^2}\right) \; dz \\=~ \frac{1}{2\pi i}\int_\mathbb{T} \frac{x}{xz- (1+z)^2} \; dz \\=~ \frac{1}{2\pi i}\int_\mathbb{T} \frac{x}{-1 + (x-2)z- z^2} \; dz$$
According to Wolframalpha I should get $\sqrt{\frac{x}{x-4}}$. Where did I commit a sin ?
I'm expecting to use the Residue theorem or Cauchy's integral formula at some point but I feel something is fishy.
 A: An ODE-based proof. If we set
$$ A(t) = \sum_{n\geq 0}\binom{2n}{n}t^n $$
we have an analytic function in a neighbourhood of the origin with radius of convergence $\rho\geq\frac{1}{4}$.
We may notice that:
$$ A'(t) = \sum_{n\geq 1}\binom{2n}{n} n t^{n-1} = \sum_{n\geq 0}\binom{2n}{n}(4n+2) t^n $$ 
while:
$$ \frac{d}{dt}\left( t A(t)\right) = \sum_{n\geq 0}\binom{2n}{n}(n+1) t^n $$
hence:
$$ A'(t) = 4\frac{d}{dt}\left( t A(t)\right) - 2A(t) = 2A(t)+4t A'(t) $$
or:
$$ \frac{A'(t)}{A(t)} = \frac{2}{1-4t}$$
that leads to:

$$ A(t) = \sum_{n\geq 0}\binom{2n}{n}t^n = \frac{1}{\sqrt{1-4t}}. \tag{1}$$

By replacing $t$ with $\frac{1}{x}$ we get:

$$ \forall x>4,\qquad \sum_{n\geq 0}\binom{2n}{n}\frac{1}{x^n}=\color{red}{\sqrt{\frac{x}{x-4}}}.\tag{2} $$

$(1)$ also follows from Lagrange's inversion theorem (that leads to Bring radical).
A: You just need to keep going:  factor the quadratic $z^2-(x-2)z+1=(z-r_1)(z-r_2)$, apply partial fractions to get
$${1\over(z-r_1)(z-r_s)}={1\over r_1-r_2}\left({1\over z-r_1}-{1\over z-r_2} \right)$$
and then use the fact that one of the roots is inside the unit circle and the other is outside when applying the Residue Theorem.  Finally, note that
$$\sqrt{x\over x-4}={x\over\sqrt{x^2-4x}}$$
