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.
Evaluate $\prod_{n=1}^\infty \frac{2^n-1}{2^n}$
6
$\begingroup$
sequences-and-series
limits
infinite-product
-
1See also [What is the value of $\prod_{i=1}^\infty 1-\frac{1}{2^i}$?](http://math.stackexchange.com/q/1200575) and other posts [linked there](http://math.stackexchange.com/questions/linked/1200575). Found [using Approach0](https://approach0.xyz/search/?q=%24%5Cprod_%7Bk%3D1%7D%5E%7B%5Cinfty%20%7D(1-%5Cfrac1%7B2%5Ek%7D)%24&p=1) – 2017-02-04
-
0@MartinSleziak Today is my first day hearing about Approach0. Are there other tools like it as well? – 2017-02-04
-
0@CodeLabMaster Basically all I am able to say about searching on this site can be found in the links [listed here](http://chat.stackexchange.com/transcript/46148/2016/10/1). If you consider Approach0 useful, you can also vote for it in [Community Promotion Ads - 2017](http://meta.math.stackexchange.com/q/25812). (And of course, you can vote there for other useful resources, too.) – 2017-02-04
-
0This 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 – 2017-02-04
-
0Here $q=1/2$ 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? – 2017-02-04
-
0I think some results are there about transcendence of theta functions in general (and also eta functions), but I need to look up the references for the same. – 2017-02-04
-
0See *Introduction to Algebraic Independence Theory* by Yuri V. Nesterenko – 2017-02-04
-
0In particular $\eta(q) $ is transcendental for all algebraic $q$ with $0<|q|<1$. And thus your particular product in question is transcendental. – 2017-02-04
1 Answers
7
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.
-
0Thanks. So, there is no other way to express the value other than that? – 2017-02-04
-
0I think this is as close to a closed form as you are going to get. One expects that the number is transcendental... – 2017-02-04
-
0I see that variants of this question have been asked before on this site, which I was not previously aware of. – 2017-02-04