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*}

Comments

Post a Comment

Popular posts from this blog

HARD INTEGRAL - PART II

Image
       Today We will compute the following integral following the same ideas of the previous post : Recall (see here ) (1) Integrating both sides of (1)  from 0 to Computing    Recall (see here ) Letting we obtain We are looking for the Imaginary part of the equation above: Computing the quantities: The Glaisher function We know that (see here ): If we integrate from 0 to x we obtain Integrating from 0 to x we obtain
Read more