Find sum of geometric-like series with binomial coefficients using complex analysis

Question

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 :)

Answer 1

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 $5-4=1$.

Answer 2

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.