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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.

John Snyder
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  • Why would you need an approximation when this sum has a closed form in terms of elliptic integrals? see https://math.stackexchange.com/questions/2613728/generating-function-of-the-sequence-binom2nn3h-n – Cave Johnson Feb 01 '18 at 18:20

1 Answers1

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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.

Jack D'Aurizio
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  • Thanks for your quick reply, this is roughly where I left it at, except I wrote it in terms of hyper geometric functions.

    To be more specific. I am essentially looking for some simple function that can approximate this for z as $z\rightarrow 1^-$.

    For example. I know that for a hypergeometric function $2F_1(a,b;a+b,z)$, $\lim{z\rightarrow 1} \frac{_2F_1(a,b;a+b,z)}{-log(1-z)} = c$. Do you think its possible to come up with anything that could do that for the RHS in (2)?

     – John Snyder Jan 29 '18 at 19:59
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    @JohnSnyder: you are already done, since $$ K(x) = \tfrac{\pi}{2}\cdot\phantom{}_2 F_1\left(\tfrac{1}{2},\tfrac{1}{2},1,x\right).$$ With extra accuracy, in a left neighbourhood of $x=1$ we have $$ K(x)\approx \frac{5}{2}\log(2)-\frac{1}{2}\log(1-x^2).$$ – Jack D'Aurizio Jan 29 '18 at 20:02
  • but I can't do anything with $K(zu)$ term, right? Since even if $z\rightarrow 1^-$, $u\in(0,1)$? – John Snyder Jan 29 '18 at 21:35
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    But you can still write the RHS of $(2)$ as $$\iint_{(0,1)^2} z\cdot K'(z(u-1)y+z),dy,du. $$ – Jack D'Aurizio Jan 29 '18 at 21:42
  • I looked it up so I am not 100% sure but just to be clear, $K'(k) = K(\sqrt{1-k^2})$? – John Snyder Jan 29 '18 at 21:53
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    @JohnSnyder: no, in the previous line $K'(x)$ simply stands for $\frac{d}{dx}K(x)$. – Jack D'Aurizio Jan 29 '18 at 21:59
  • Let us continue this discussion in chat. – John Snyder Jan 29 '18 at 22:29
  • I don't quite follow, can your double integral be used to infer about the behavior of the RHS of $(2)$ in a left neighborhood of $z=1$?

    Alternatively, I was thinking that with respect to $u$, the integrand in the RHS of $(2)$ would be well behaved for $u \in (0,1-\varepsilon)$, right? Therefore, I could determine the behavior of the RHS of $(2)$ by splitting up the integral into $u \in (0,1-\varepsilon)$ and $u \in (1-\varepsilon,1)$ and using the limiting behavior of $K(z)$ or $_2F_1(\frac{1}{2},\frac{1}{2};1;z)$ for the second part.

    Is this reasonable?

    Thanks.

     – John Snyder Jan 30 '18 at 17:34