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Asymptotic expansions of truncated hypergeometric series for $1/\pi$

Number Theory 2024-07-24 v2

Abstract

In this paper, we consider rational hypergeometric series of the form pπ=k=0ukwithuk=(12)k(q)k(1q)k(k!)3(r+sk)tk,\frac{p}{\pi}= \sum_{k=0}^\infty u_k\quad\text{with}\quad u_k=\frac{\left(\frac{1}{2}\right)_k \left(q\right)_k \left(1-q\right)_k}{(k!)^3}(r+s\,k)\,t^k, where (a)k(a)_k denotes the Pochhammer symbol and p,q,r,s,tp,q,r,s,t are algebraic coefficients. Using only the first n+1n+1 terms of this series, we define the remainder Rn=pπk=0nuk=k=n+1uk.\mathcal{R}_n = \frac{p}{\pi} - \sum_{k=0}^n u_k=\sum_{k=n+1}^\infty u_k. We consider an asymptotic expansion of Rn\mathcal{R}_n. More precisely, we provide a recursive relation for determining the coefficients cjc_j such that Rn=(12)n(q)n(1q)nn!3ntn(j=0J1cjnj+O(nJ)),n. \mathcal{R}_n = \frac{\left(\frac{1}{2}\right)_n \left(q\right)_n \left(1-q\right)_n}{n!^3}nt^n\left(\sum_{j=0}^{J-1}\frac{c_j}{n^j}+\mathcal{O}\left(n^{-J}\right)\right),\qquad n \rightarrow \infty. Here we need J<J<\infty to approximate Rn\mathcal{R}_n, because (like the Stirling series) this series diverges if JJ\rightarrow\infty. By applying our recursive relation to the Chudnovsky formula, we solve an open problem posed by Han and Chen.

Keywords

Cite

@article{arxiv.2401.05419,
  title  = {Asymptotic expansions of truncated hypergeometric series for $1/\pi$},
  author = {Lorenz Milla and Chao-Ping Chen},
  journal= {arXiv preprint arXiv:2401.05419},
  year   = {2024}
}

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12 pages