(Too long for a comment.)
There's a (somewhat) simpler (Eulerian) continued fraction:
$$\sum_{k=1}^\infty \frac1{k^s}=\cfrac1{1-\cfrac{\left(1-\frac12\right)^s}{1+\left(1-\frac12\right)^s-\cfrac{\left(1-\frac13\right)^s}{1+\left(1-\frac13\right)^s-\cfrac{\left(1-\frac14\right)^s}{1+\left(1-\frac14\right)^s-\cdots}}}}$$frac1{k^s}=\sum_{k=1}^{\infty} \prod_{j=2}^k \left(1-\frac1{j}\right)^s=\cfrac1{1-\cfrac{\left(1-\frac12\right)^s}{1+\left(1-\frac12\right)^s-\cfrac{\left(1-\frac13\right)^s}{1+\left(1-\frac13\right)^s-\cfrac{\left(1-\frac14\right)^s}{1+\left(1-\frac14\right)^s-\cdots}}}}$$
but as you can see from comparing successive convergents of this continued fraction and the successive partial sums of the Dirichlet series, it's not terribly useful.
Also,
$$e^{-2\mathrm{arcoth}((2k+1)z)}=1-\frac2{1+(2k+1)z}$$$e^{-2z\,\mathrm{arcoth}(2k+1)}=\left(\frac{k}{k+1}\right)^z$$
so your CF could certainly be simplified a fair bit...
