Ramanujan´s Psi Sum

In this blog entry we will prove Ramanujan´s Psi Sum, namely


\begin{align*} \sum_{n=-\infty}^{\infty} \frac{(a;q)_n}{(b;q)_n} x^n=\frac{(a x;q)_{\infty}(q / a x;q)_{\infty}(q;q)_{\infty}(b / a;q)_{\infty}}{(x;q)_{\infty}(b / a x;q)_{\infty}(b;q)_{\infty}(q / a;q)_{\infty}} \end{align*}


Click here for the proof.


As corollaries of the above formula we show that


\begin{align*} \left(\sum_{n=-\infty}^\infty q^{n^2} \right)^2=2 \sum_{n=-\infty}^\infty \frac{q^n}{1+q^{2n}} \end{align*}



\begin{align*} \left(\sum_{n=-\infty}^\infty q^{n^2} \right)^4=1+8 \sum_{n=-\infty}^\infty \frac{n q^n}{1+(-q)^{n}} \end{align*}


wich relies also on Jacobi triple product proved here:


\begin{align*} \sum_{n=-\infty}^{\infty} q^{n^{2}} z^{n}=\prod_{n=1}^{\infty}\left(1+z q^{2 n-1}\right)\left(1+z^{-1} q^{2 n-1}\right)\left(1-q^{2 n}\right) \end{align*}

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