Let $w_k$ be a primitive *k* th root of unity, where *k* is a power of 2. In response to a question, Robert Israel gave the solution,

$$\sum_{n=0}^\infty \frac{(-1)^n}{\binom{kn}{kn/2}} = \frac{2^k}{2^k + 1} + \frac{1}{k} \sum_{j=0}^{k-1} \frac{\omega_k^{j+1/2} \arcsin(\omega_k^{j+1/2}/2)}{2 (1- \omega_k^{2j+1}/4)^{3/2}}$$

The sum is real, but one can see the terms are complex. Renzo Sprugnoli found a version for *k* = {2,4} in *real* terms as,

$\sum_{n=0}^\infty \frac{(-1)^n}{\binom {2n}n} = \frac{4}{5} - \frac{4\sqrt{5}}{25}\ln\left(\frac{1+\sqrt{5}}{2}\right)$

$\sum_{n=0}^\infty \frac{(-1)^n}{\binom {4n}{2n}} = \frac{16}{17}+\frac{4\sqrt{34}(-2+\sqrt{17})}{289\sqrt{-1+\sqrt{17}}}\arctan\left(\frac{\sqrt{2}}{\sqrt{-1+\sqrt{17}}}\right) -\frac{2\sqrt{34}(2+\sqrt{17})}{289\sqrt{1+\sqrt{17}}} \ln\left(\frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\right )$

I found that, given the *Dedekind eta function* $\eta(z)$ and,

$t_1 = \frac{1+\sqrt{-5}}{2}$

$t_2 = \frac{1+\sqrt{-17}}{2}$

$\zeta_{48} = \exp(\pi \rm i/24)$

then,

$$\frac{1}{2} \left(\frac{\zeta_{48}\eta(t_1)}{\eta(2t_1)}\right)^4 = \frac{1+\sqrt{5}}{2}$$

$$\frac{1}{2} \left(\frac{\zeta_{48}\eta(t_2)}{\eta(2t_2)}\right)^4 = \frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}$$

*Question*: Is this coincidence? For the next case *k* = 8, how do we convert Israel's complex terms into *real ones* (akin to Sprugnoli's) to check if the argument of the natural log can be similarly expressed by an eta quotient?