# Adventure with infinite series, a curiosity

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.

• I don't see how $3$ fits the pattern in $3,5,10,20,...$. Oct 8, 2018 at 18:18

It is moderately nice, I would say. We have $$\sum \binom{2k}k x^k=(1-4x)^{-1/2}$$ for $$|x|<1/4$$. If we need only terms with $$k$$ divisible by 4 and $$x=2^{-5/2}$$, $$4x=2^{-1/2}$$, we get $$\sum \binom{8k}{4k}2^{-10k}=\frac14\sum_{w^4=1}(1-w/\sqrt{2})^{-1/2}$$ and so on.

More generally, it seems that $$\sum_{k\ge0}\binom{2^{j+1}k}{2^jk}2^{-a_{j+2} k}=2^{-j}\sum_{w^{2^j}=1}(1-w/\sqrt2)^{-1/2}$$ where $$a_1=2,a_2=3,a_{j+1}=a_j+a_{j-1}+\dots+a_1$$ (cf. A257113).

• Note: this has been confirmed up to $j=7$. Oct 8, 2018 at 16:34

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.

• Well I would say under the "and so on" at the end of another answer is most probably understood (more or less an equivalent of) this more general formula. Jan 24, 2019 at 18:14
• @მამუკაჯიბლაძე Yes, you are right. The above is just a very usual and basic technique.
– Zhou
Jan 25, 2019 at 1:15

Mathematica says, for the first:

$$\, _4F_3\left(\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8};\frac{1}{4},\frac{1}{2},\frac{ 3}{4};\frac{1}{4}\right)$$

and

$$\, _8F_7\left(\frac{1}{16},\frac{3}{16},\frac{5}{16},\frac{7}{16},\frac{9}{16},\frac{11}{16} ,\frac{13}{16},\frac{15}{16};\frac{1}{8},\frac{1}{4},\frac{3}{8},\frac{1}{2},\frac{5}{8}, \frac{3}{4},\frac{7}{8};\frac{1}{16}\right)$$ for the second.

• What's with the downvotes? Oct 9, 2018 at 0:36
• Is the answer wrong? Oct 9, 2018 at 3:25
• It's not wrong, but I think this is obvious from the definition of the generalized hypergeometric function ${}_pF_q$, and doesn't seem to be a helpful step toward a closed form. (I didn't downvote, fwiw.) Oct 9, 2018 at 5:58
• @DavidZhang Interestingly, for the first two cases, mathematica gives non-hypereometric forms. Oct 9, 2018 at 16:25