Ratio of hypergeometric function with a "Harmonic twist" to a hypergeometric function Continuing with a problem I am working which involves the work here, I am faced with the following expression.
\begin{equation}
\frac{1}{2\,_2F_1\left(\frac{1}{2},\frac{1}{2};1;z\right)}
\sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n}{\left(1\right)_n}\frac{z^n}{n!}\left[H_{n-1/2} - 2H_n\right]
\end{equation}
where  $H_{n-1/2}=\sum_{k=1}^{n}\frac{2}{2k-1}$ and $H_n=\sum_{k=1}^{n}\frac{1}{k}$.
I recognize that the series is almost a hypergeometric series, but I would like to approximate this expression to obtain a function of $z$.  Any help is greatly appreciated.
 A: First things first: $\phantom{}_2 F_1\left(\tfrac{1}{2},\tfrac{1}{2};1;m\right)=\tfrac{2}{\pi}\,K(m)$ and the complete elliptic integral of the first kind is very simple to approximate through the AGM mean (see Borwein et al.). It just remains to estimate
$$ \sum_{n\geq 0}\frac{\binom{2n}{n}^2 z^n}{16^n}\sum_{k=1}^{n}\left(\frac{2}{2k-1}-\frac{1}{k}\right)=2\log 2\,K(z)+\sum_{n\geq 0}\frac{\binom{2n}{n}^2 z^n}{16^n}\left(-2\int_{0}^{1}\frac{t^{2n}}{1+t}\,dt\right) $$
which equals
$$ 2\log 2\,K(z)-2\int_{0}^{1}\sum_{n\geq 0}\frac{\binom{2n}{n}^2 (zt^2)^n}{16^n}\cdot\frac{dt}{1+t}=2\log 2\,K(z)-\frac{4}{\pi}\int_{0}^{1}\frac{K(zt^2)}{1+t}\,dt$$
or
$$ 2\log 2\,K(z)-\frac{2}{\pi}\int_{0}^{z}\frac{K(t)}{t+\sqrt{tz}}\,dt $$
and plenty of accurate approximations for $K(t)$ (for $t$ in a right neighbourhood of the origin) are known, so the numerical evaluation of the given "monster" is actually pretty simple with the right tools. $$\sum_{n\geq 0}\frac{\binom{2n}{n}^2 z^n}{16^n}H_n$$
can be managed in a similar way.
