I am following up on a post I made a couple months ago as I am revisiting this problem. I desire a way to approximate the sum
$$\sum_{n\geq 0}\frac{\binom{2n}{n}^2 z^n}{16^n}H_n$$
for a specified value of $z$. So far, I have tried noting that
$$H_n = \int_0^1 \frac{1-t^n}{1-t} dt$$ and distributing the terms of the sum into the integral and working with the hypergeometric function. However I am having trouble proceeding from there. Any help would be appreciated.
According to Mathematica's notation (where the argument of the complete elliptic integral of the first kind $K(x)$ is regarded as the elliptic modulus) we have $$ \sum_{n\geq 0}\frac{\binom{2n}{n}^2}{16^n}\,u^n = \frac{2}{\pi}\,K(u)\tag{1}$$ hence it is simple to check that $$ \sum_{n\geq 0}\frac{\binom{2n}{n}^2 H_n}{16^n}\,z^n =\frac{2}{\pi}\int_{0}^{1}\frac{K(zu)-K(z)}{u-1}\,du\tag{2}$$ and the integral in the RHS of $(2)$ can be numerically approximated with arbitrary precision by exploiting the relation between $K$ and the $\text{AGM}$ mean. For instance, $K(x)\approx \frac{\pi}{1+\sqrt{1-x}}$ or $K(x)\approx\frac{\pi}{2(1-x)^{1/4}}$. Similar series appear in the computation of $\int_{0}^{1}K(x)\log(1-x)\,dx$ and $\int_{0}^{1}E(x)\log(1-x)\,dx$ through Fourier-Legendre series expansions.