Evaluate $\prod_{n=1}^\infty \frac{2^n-1}{2^n}$

Question

Is there a closed form expression for this limit? $$\prod_{n=1}^\infty \frac{2^n-1}{2^n}$$ Wolfram Alpha says $0.2887880950866024212788997219292307800889\dots$ and the Inverse Symbol Calculator found nothing but the above expression.

Answer 1

It is equal to $\phi(1/2)$ where $\phi(q)$ is the Euler function, defined by $$ \phi(q)=\prod_{n=1}^{\infty}(1-q^n). $$ This is closely related to the $q$-Pochhammer symbol as well.

From Euler's pentagonal number theorem one obtains the following rapidly convergent binary expansion for $\phi(1/2)$: $$ \phi(1/2)=\sum_{n=-\infty}^{\infty}(-1)^n2^{(-3n^2+n)/2}, $$ that is, $$ \phi(1/2)=1-\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^5}+\frac{1}{2^7}-\frac{1}{2^{12}}-\frac{1}{2^{15}}+\cdots $$ with the signs repeating in the pattern $-,-,+,+$ and the exponents growing quadratically.

Proof of transcendence.

As pointed out by P. Singh in the comments above, $\phi(1/2)$ is known to be transcendental. This follows from results established in

Nesterenko, Yu. V. (1996), Modular functions and transcendence questions, Mat. Sb. p. 65-96 MR1422383.

Since this article does not have open access, I am posting the statement of the main theorem below.

We use the following identity (whose proof is indicated below) $$ \phi(q)^{24}=\frac{Q(q)^3-R(q)^2}{1728q} $$ to observe that, if $\phi(1/2)$ was algebraic, then $Q(1/2)$ and $R(1/2)$ would be algebraically dependent, contradicting the theorem. Thus $\phi(1/2)$ is transcendental, as claimed.

Proof of the identity: This is equivalent to a well-known identity expressing the modular discriminant in terms of Eisenstein series.