<s>[INST] 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.
 [/INST] 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
$$\sqrt{x\over x-4}={x\over\sqrt{x^2-4x}}$$
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