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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.

John Snyder
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1 Answers1

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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.

Jack D'Aurizio
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  • At the risk of asking a low level question, why is \begin{equation} \sum_{k=1}^{n}\left(\frac{2}{2k-1} - \frac{1}{k}\right) = -2\int_{0}^{1}\frac{t^{2n}}{1+t}dt? \end{equation} In an effort to understand I have evaluated them numerically in software as well as tried to manipulate the integral representations of harmonic series and I am not seeing it. Additionally, my term is $H_{n-1/2} - 2H_n$, would that change things or can the same technique be used? Thanks for your time. – John Snyder Nov 01 '17 at 12:54
  • I cannot seem to verify the above equality analytically or numerically for $H_{n-1/2} - H_{n}$ or my quantity of $H_{n-1/2} - 2H_{n}$. Do you have any suggestions? – John Snyder Nov 02 '17 at 18:06
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    $$\frac{t^{2n}}{1+t}=\frac{t^{2n}-t^{2n+1}}{1-t^2}$$ expand $\frac{1}{1-t^2}$ as a geometric series and perform termwise integration. – Jack D'Aurizio Nov 02 '17 at 18:09
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    There is a little typo, however. We have $$\sum_{k=1}^{n}\left(\frac{2}{2k-1}-\frac{1}{k}\right)=H_{n-1/2}-H_n+2\log 2 = 2\log 2-2\int_{0}^{1}\frac{t^{2n}}{1+t},dt $$ – Jack D'Aurizio Nov 02 '17 at 18:12
  • Thank you, I will keep working through this. Additionally, in the first line of your initial answer, I believe you meant $\phantom{}_2 F_1\left(\tfrac{1}{2},\tfrac{1}{2};1;m^2\right)=\tfrac{2}{\pi},K(m)$ – John Snyder Nov 02 '17 at 18:25
  • @JohnSnyder: that is just a notation issue. I used $K(m)$ meaning that $m$ is the elliptic modulus (Mathematica's notation) and not the elliptic parameter. – Jack D'Aurizio Nov 02 '17 at 18:31
  • I am revisiting this and am having some trouble so I posted a followup to this – John Snyder Jan 29 '18 at 19:10