TITLE: On sums such as $\sum_{k=0}^\infty \binom{2k}{k}\frac{1}{8^k}=\sqrt{2}$
QUESTION [8 upvotes]: This identity is a special case of a more general formula found here, and known at least since 1972 (published in Abramowitz and Stegun 1972, p. 555.) Many remarkable series involving the inverse of binomial coefficients are known and listed in the same source, see also here for a way to derive them.
However, there are plenty of interesting series involving binomial coefficients (not their inverse) that are not listed in the references that I checked. For instance:
$$\sum_{k=0}^\infty \binom{3k}{k}\frac{1}{8^k}=\frac{4\sqrt{10}}{5}\cos\Big(\frac{1}{3}\arcsin \frac{3\sqrt{6}}{8}\Big)$$
$$\sum_{k=0}^\infty \binom{4k}{2k}\frac{1}{32^k}= \sin\frac{\pi}{8}+\cos\frac{\pi}{8}$$
Another example is:

Question
How do you prove these results? I found them using WolframAlpha symbolic calculator, see here for an example. One of these results is proved here, but I am looking for a proof that would apply to a broad class of such series.  
Background
The reason that I am interested in powers of two ($8^{-k}, 32^{-k}$) is because I am looking for series converging to irrational numbers, with each term being a fraction: the denominator is a power of two, and the numerator is an integer (a binomial coefficient in this case.) The goal is to gain some insights in the binary digit distribution of numbers such as $\sqrt{2}$.
For instance, a result that could be useful for me is the following:
$$\sqrt{2} = \lim_{n\rightarrow\infty}\frac{P_n}{8^n} = \lim_{n\rightarrow\infty}\frac{1}{8^n}\sum_{k=0}^n 8^{n-k}\binom{2k}{k},$$
with $P_n$ being an integer.

REPLY [3 votes]: Another way is to use the integral representation of the binomial coefficient $$\dbinom{n}{k}=\frac{1}{2\pi i}\oint_{|z|=1}\frac{\left(1+z\right)^{n}}{z^{k+1}}dz.$$ For example we have $$\sum_{k\geq0}\frac{1}{8^{k}}\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{2k}}{z^{k+1}}dz=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{1}{z}\sum_{k\geq0}\frac{1}{8^{k}}\frac{\left(1+z\right)^{2k}}{z^{k}}dz$$ $$=-\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{8}{z^{2}-6z+1}dz$$ and since we have a pole at $z=3-2\sqrt{2}$ we get $$\sum_{k\geq0}\frac{1}{8^{k}}\dbinom{2k}{k}=\color{red}{\sqrt{2}}.$$ In a similar way $$\sum_{k\geq0}\frac{1}{8^{k}}\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{3k}}{z^{k+1}}dz=-\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{8}{z^{3}+3z^{2}-5z+1}dz$$ and since we have a pole at $z=\sqrt{5}-2$ we get $$\sum_{k\geq0}\frac{1}{8^{k}}\dbinom{3k}{k}=\color{red}{1+\frac{3}{\sqrt{5}}}.$$