Find sum of geometric-like series with binomial coefficients using complex analysis Studying analytic number theory, I stumbled across the problem of finding the sum of the series
$\sum_{n=0}^{\infty}\binom{2n}{n}\left(\frac{1}{5}\right)^n$
A professor gave me the hint of "using basic complex analysis" but honestly, I haven't been able to get anywhere. Appreciate any help/comments :)

Following the hints we introduce
$${2n\choose n} =
\frac{1}{2\pi i} \int_{|z|=\varepsilon} \frac{(1+z)^{2n}}{z^{n+1}} \; dz.$$
We get for the sum
$$\frac{1}{2\pi i} \int_{|z|=\varepsilon} \frac{1}{z}
\sum_{n\ge 0} \frac{(1+z)^{2n}}{5^n\times z^{n}} \; dz.$$
We must now determine $\varepsilon$ for the geometric series to converge.
We need
$$|(1+z)^2| \lt 5 |z|.$$
Now $$|(1+z)^2| \le (1+\varepsilon)^2$$ so we have an admissible
$\varepsilon$ if
$$(1+\varepsilon)^2 \lt 5\varepsilon.$$
which is
$$1-3\varepsilon+\varepsilon^2 \lt 0.$$
The roots are           
$$\rho_{0,1} = \frac{3\mp\sqrt{5}}{2}.$$
($\rho_0$ is the smaller of the two)
Then $(\varepsilon-\rho_0) (\varepsilon-\rho_1) \lt 0$ if
$$\rho_0 \lt \varepsilon \lt \rho_1.$$
so we have convergence in an annulus delimited by two circles of radius
$\rho_0 \lt \rho_1.$
We sum the series and get
$$-5 \times \frac{1}{2\pi i} \int_{|z|=\varepsilon} 
\frac{1}{z^2-3z+1} \; dz.$$
The only pole inside the circle is the one at $\rho_0$ and we find
$$-5 \times \mathrm{Res}_{z=\rho_0} \frac{1}{z^2-3z+1}
= -5 \frac{1}{2\rho_0-3} = -5 \frac{1}{3-\sqrt{5}-3} = \sqrt{5}.$$
This is our answer.
