Jimmy.McLaughlinAffiliation: West Chester University PA Title Of Talk: $q$-extensions of formulae for $\\pi$ with free parameters Abstract: We prove several $q$-series identities which have one or more free parameters on one side. An example is given by \begin{multline} \sum_{n=1}^{\infty} (q^4;q^6)_{n}(q^6;q^6)_{n-1}q^{n} \frac{(q^5/bc;q^6)_n}{(q^5/b,q^5/c,q^6;q^6)_n} \\ - \sum_{n=1}^{\infty}\frac{(q^4;q^6)_{n}(q^6;q^6)_{n-1}} {(q^5 ,q;q^6)_n}q^{n} \frac{(1- q^{12n-1})(1/q,b,c;q^6)_n} {(1-1/q)(q^5/b,q^5/c,q^6;q^6)_n}\left( \frac{-q^5}{bc}\right)^n q^{3(n^2-n)}\\ = \frac{1}{3}\left( \frac{\psi^3(q)}{\psi(q^3)}-1 \right). \end{multline} Upon letting $q \to 1$ on both sides leads to Ramanujan-type series for expressions involving $\pi$. From the identity above, for example, one gets that if $b$ and $c$ are complex numbers such that $Re(b+c)<13/2$ and neither is a positive integer $\equiv 5 \pmod{6}$, then \begin{multline} \sum_{n=1}^{\infty}\frac{\left(\frac{2}{3}\right)_n \left(\frac{1}{6} (-b-c+5)\right)_n}{ n \left(\frac{5-b}{6}\right)_n \left(\frac{5-c}{6}\right)_n} - \sum_{n=1}^{\infty}\frac{(12 n-1) \left(\frac{2}{3}\right)_n \left(\frac{b}{6}\right)_n \left(\frac{c}{6}\right)_n(-1)^n }{ n(6n-1) \left(\frac{1}{6}\right)_n \left(\frac{5-b}{6}\right)_n \left(\frac{5-c}{6}\right)_n}\\ =\pi \sqrt{3}. \end{multline}
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Last update made Tue Mar 10 21:28:04 CDT 2026.
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