TITLE: Adventure with infinite series, a curiosity
QUESTION [10 upvotes]: It is easily verifiable that
$$\sum_{k\geq0}\binom{2k}k\frac1{2^{3k}}=\sqrt{2}.$$
It is not that difficult to get
$$\sum_{k\geq0}\binom{4k}{2k}\frac1{2^{5k}}=\frac{\sqrt{2-\sqrt2}+\sqrt{2+\sqrt2}}2.$$

Question. Is there something similarly "nice" in computing
  $$\sum_{k\geq0}\binom{8k}{4k}\frac1{2^{10k}}=?$$
  Perhaps the same question about
  $$\sum_{k\geq0}\binom{16k}{8k}\frac1{2^{20k}}=?$$

NOTE. The powers of $2$ are selected with a hope (suspicion) for some pattern.

REPLY [3 votes]: Let $$f(x):=\sum_{k\ge 0}\binom{2k}{k}x^k=\frac{1}{\sqrt{1-4x}}$$
with $|x|<1/4$. We have for $N,j\in\mathbb{Z}_+$,
\begin{align}
\frac{1}{N}\sum_{r=1}^Ne(-{jr}/{N})f\left(xe({r}/{N})\right)&=\frac{1}{N}\sum_{r=1}^Ne(-{jr}/{N})\sum_{k\ge 0}\binom{2k}{k}x^ke({kr}/{N})\\
&=\sum_{k\ge 0}\binom{2k}{k}x^k\frac{1}{N}\sum_{r=1}^Ne({(k-j)r}/{N}),
\end{align}
where $e(x):=e^{2\pi{\rm i}x}$. Hence clearly,
$$\sum_{\substack{k\ge 0\\ k\equiv j\pmod N}}\binom{2k}{k}x^k=\frac{1}{N}\sum_{r=1}^N\frac{e(-{jr}/{N})}{\sqrt{1-4xe(r/N)}},$$
holds for all $N,j\in\mathbb{Z}_+$, which is a more general result.