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

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.

• See also What is the value of $\prod_{i=1}^\infty 1-\frac{1}{2^i}$? and other posts linked there. Found using Approach0 – Martin Sleziak Feb 4 '17 at 8:38
• @MartinSleziak Today is my first day hearing about Approach0. Are there other tools like it as well? – CodeLabMaster Feb 4 '17 at 9:02
• @CodeLabMaster Basically all I am able to say about searching on this site can be found in the links listed here. If you consider Approach0 useful, you can also vote for it in Community Promotion Ads - 2017. (And of course, you can vote there for other useful resources, too.) – Martin Sleziak Feb 4 '17 at 9:06
• This is essentially the Dedekind's eta function $$\eta(q) =q^{1/24}\prod_{n=1}^{\infty}(1-q^{n})$$ and it has closed form representation in terms of complete elliptic integrals of the first kind if $q=e^{-\pi\sqrt{r}}$ where $r$ is a positive rational number – Paramanand Singh Feb 4 '17 at 9:38
• Here $q=1/2$ so this is not the case. By the way, is it known whether special values of the eta function are transcendental, for example when $q$ is rational and not equal to 0 or 1? – pre-kidney Feb 4 '17 at 9:46

## 1 Answer

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.

• Thanks. So, there is no other way to express the value other than that? – Kay K. Feb 4 '17 at 6:46
• I think this is as close to a closed form as you are going to get. One expects that the number is transcendental... – pre-kidney Feb 4 '17 at 6:48
• I see that variants of this question have been asked before on this site, which I was not previously aware of. – pre-kidney Feb 4 '17 at 9:47