<s>[INST] 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 :)
 [/INST] Consider the integral
$$\int_0^{2\pi}4^n\cos^{2n}(x)\:dx$$
Expanding out using Euler's formula and binomial expansion, we get that
$$\int_0^{2\pi}4^n\cos^{2n}(x)\:dx = \int_0^{2\pi}(e^{ix}+e^{-ix})^{2n}\:dx = \sum_{k=0}^{2n} {2n \choose k} \int_0^{2\pi}e^{i(2n-2k)x}\:dx$$
The integral on the right will always be $0$ unless the exponent is $0$, meaning the only term that survives the summation is $k=n$
$$\sum_{k=0}^{2n} {2n \choose k} \int_0^{2\pi}e^{i(2n-2k)x}\:dx = {2n \choose n}\cdot 2\pi$$
Now going backwards, we can substitute this value into the summation:
$$\sum_{n=0}^\infty {2n \choose n}\left(\frac{1}{5}\right)^n = \frac{1}{2\pi}\int_0^{2\pi} \:dx \sum_{n=0}^\infty \left(\frac{4}{5}\cos^2(x)\right)^n = \frac{1}{2\pi}\int_0^{2\pi} \frac{1}{1-\frac{4}{5}\cos^2(x)}\:dx $$
Then rationalize and use trig identities to get the following expression:
$$= \frac{5}{2\pi}\int_0^{2\pi} \frac{1}{\cos^2(x) + 5\sin^2(x)}\:dx = \frac{\sqrt{5}}{2\pi}\int_0^{2\pi} \frac{\sqrt{5}\sec^2(x)}{1 + 5\tan^2(x)}\:dx$$
$$= \frac{\sqrt{5}}{2\pi}\tan^{-1}\left(\sqrt{5}\tan(x)\right)\Biggr|_0^{2\pi} = \frac{\sqrt{5}}{2\pi}\left(\frac{\pi}{2} + \pi + \frac{\pi}{2}\right) = \sqrt{5}$$
Leaving us with our final result
$$\sum_{n=0}^\infty {2n \choose n}\left(\frac{1}{5}\right)^n = \sqrt{5}$$
This only worked out so nicely because $5$ is the only number with the special property that $5-4=1$.
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