4

The source of idea is from here

Where $$\psi(q)=\sum_{k=0}^{\infty}q^{\frac{k(k+1)}{2}}$$

Prove that,

(1)

$$\prod_{k=1}^{\infty}\left(1-e^{-\pi{k\sqrt n}}\right)=\frac{\sqrt2 g_n^2\psi(e^{-\pi\sqrt n})}{e^{\frac{\pi\sqrt n}{12}}}$$

On my recent post here Paramananad Singh showed that our question is a simple case of Ramanujan $G_n$ and $g_n$ Function. We found another similiar Ramanujan-type infinite product and ask if anyone can help us to prove (1).

We are going to verify (1) numerically

We used the values $g_n$ from here

$g_{10}=\sqrt{\frac{1+\sqrt5}{2}}$

$$\prod_{k=1}^{\infty}\left(1-e^{-\pi{k\sqrt{10}}}\right)=\frac{\phi\sqrt2 \psi(e^{-\pi\sqrt{10}})}{e^{\frac{\pi\sqrt{10}}{12}}}=0.9999515288...$$

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1 Answers1

1

Let $q = e^{-\pi\sqrt{n}}$ and then we need to evaluate the product $$F(q) = \prod_{k = 1}^{\infty}(1 - q^{k})\tag{1}$$ The Ramanujan class invariant $g_{n}$ is given by $$g_{n} = 2^{-1/4}q^{-1/24}\prod_{k = 1}^{\infty}(1 - q^{2k - 1})\tag{2}$$ Next we discuss the Ramanujan theta function $$\psi(q) = \sum_{k = 0}^{\infty}q^{k(k + 1)/2} = \prod_{k = 1}^{\infty}\frac{1 - q^{2k}}{1 - q^{2k - 1}} = \prod_{k = 1}^{\infty}\frac{1 - q^{k}}{(1 - q^{2k - 1})^{2}}\tag{3}$$ From the above equations it is clear that $$\psi(q) = \frac{F(q)}{2^{1/2}q^{-1/12}g_{n}^{2}}$$ and thus $$F(q) = \prod_{k = 1}^{\infty}(1 - e^{-k\pi\sqrt{n}}) = \frac{\sqrt{2}g_{n}^{2}\psi(e^{-\pi\sqrt{n}})}{e^{\pi\sqrt{n}/12}}$$

The equality between series and product representation of $\psi(q)$ in $(3)$ is proved via the use of Jacobi Triple Product identity (original proof of this identity by Jacobi is available here) $$\sum_{k = -\infty}^{\infty}z^{k}q^{k^{2}} = \prod_{k = 1}^{\infty}(1 - q^{2k})(1 + zq^{2k - 1})(1 + z^{-1}q^{2k - 1})\tag{4}$$ by replacing $q$ with $q^{2}$ and $z = q$. With this replacement LHS of $(4)$ is the series representation of $\psi(q)$ and RHS of $(4)$ is the product representation of $\psi(q)$.

Paramanand Singh
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