# Is the infinite product $\prod_{i=0}^{\infty}(1+\frac{1}{2^{3^i}})$ transcendental?

Is the following number algebraic or transcendental? $$P:=\prod_{i=0}^{\infty}\left(1+\frac{1}{2^{3^i}}\right)$$ We could also define it as follows: let A be the set of natural numbers which contain only 0's and 1's in base 3. Then our number is $$\sum_{a\in A}2^{-a}.$$ I have found a proof that $$P$$ is irrational (which is a partial result on this question): for $$n\geq1$$, let $$P_n:=\prod_{i=0}^n(1+\frac{1}{2^{3^i}})$$. Then $$P_n$$ is a rational number with denominator $$D_n=2^{3^0+3^1+\dots+3^n}$$. Now we have $$0<|P_n-P|=\sum_{i\in A\mid i\geq3^{n+1}}2^{-i}<\sum_{i=3^{n+1}}^{\infty}2^{-i}=\frac{1}{2^{3^{n+1}-1}}=\frac{1}{D_n^2}.$$ This holds for all $$n$$, and we have $$D_n\rightarrow\infty$$ as $$n\rightarrow\infty$$. Therefore, $$P$$ has irrationality measure at least $$2$$, which implies that $$P$$ is irrational.

• Every real number has irrationality measure at least $2$, I think. What you need to show is that the irrationality measure is greater than $2$ (this would also show that $P$ is transcendent, btw). And the decomposition above isn’t going to work. But here, you can see that the base two expansion of $P$ is nonperiodic – thus the number is irrational. Jun 29, 2021 at 8:48
• According to Wikipedia, rational numbers have irrationality measure equal to 1: en.wikipedia.org/wiki/Liouville_number#Irrationality_measure Jun 29, 2021 at 8:53
• There was a mistake in my proof; I fixed it. You are right, you can immediately show that the base two expansion is nonperiodic, so now we have two proofs that P is irrational. Jun 29, 2021 at 8:57
• @Mindlack. "The binary expansion of $P$ is non-periodic" does it. We can also show that if $P\not\in\Bbb Q$ then $P$ is a Liouville number...BTW, using elementary properties of Farey sequences we can show that if $x\in\Bbb R\setminus \Bbb Q$ there exist infinitely many $(a,b)\in \Bbb Z\times \Bbb Z^+$ such that $|x-a/b|<1/(b^2\sqrt 5)$. We can't replace the $\sqrt 5$ with anything larger if $x$ is the Golden Ratio. A much deeper result, for which Roth won a Fields Medal, is that if an irrational real $x$ has irrationality measure $>2$ then $x$ is transcendental.(As you said). Jun 29, 2021 at 10:08
• $P_n$ has denominator $2^{3^n}=LCM ( \{2^{3^i}: 0\le i\le n\})$ Jun 29, 2021 at 10:14

Using a technique called Mahler's transcendence method, we can prove that $$\prod_{i=0}^{\infty}\left(1+\frac{1}{2^{3^i}}\right)$$ is transcendental. We shall use the following statement from the beginning of this article:

Let $$p\in\mathbb Z_{\geq2}$$. Let $$f(z)$$ be a transcendental function which is holomorphic on the open unit disk and which satisfies the functional equation $$f(z) = a(z) f(z^p) + b(z)$$, where $$a(z)$$ and $$b(z)$$ are polynomials with algebraic coefficients, and for which $$f(0)$$ is algebraic. If $$\alpha$$ is an algebraic number, $$0<|\alpha| < 1$$, and if $$a(\alpha^{p^k})\neq0$$ for all $$k\in\mathbb Z_{\geq0}$$, then $$f(\alpha)$$ is transcendental.

We let $$p:=3$$ and define $$f:D\rightarrow\mathbb C$$ by $$f(z):=\prod_{i=0}^{\infty}\left(1+z^{3^i}\right)$$ for all $$z$$ in the open unit disk $$D$$. Then $$f$$ is a holomorphic function which satisfies the functional equation $$f(z)=(1+z)f(z^3)$$. Also, $$f(0)=1$$ is algebraic. However, before we can apply the theorem, we need to show that $$f(z)$$ is a transcendental function. For this we follow the strategy described in Wadim Zudilin's answer to this MathOverflow question.

We shall prove by contradiction that $$f(z)$$ is transcendental over $$\mathbb C(z)$$. This is a stronger statement than we need, but I don't know a shorter proof which proofs the weaker statement that $$f(z)$$ is a transcendental function. So suppose that $$f(z)$$ is algebraic over $$\mathbb C(z)$$. Then there is a unique irreducible equation $$$$f(z)^n+a_{n-1}(z)f(z)^{n-1}+a_{n-2}(z)f(z)^{n-2}+\dots+a_1(z)f(z)^1+a_0(z)=0$$$$ with coefficients $$a_0,\dots,a_{n-1}\in \mathbb C(z)$$. Substituting $$z^3$$ for $$z$$ we obtain that $$f(z^3)^n+a_{n-1}(z^3)f(z^3)^{n-1}+a_{n-2}(z^3)f(z^3)^{n-2}+\dots+a_1(z^3)f(z^3)^1+a_0(z^3)=0,$$ and using our functional equation $$f(z^3)=\frac{1}{z+1}f(z)$$ we obtain that $$\frac{1}{(z+1)^n}f(z)^n+\frac{1}{(z+1)^{n-1}}a_{n-1}(z^3)f(z)^{n-1}+\dots+\frac{1}{z+1}a_1(z^3)f(z)^1+a_0(z^3)=0.$$ This gives us that $$$$f(z)^n+(z+1)a_{n-1}(z^3)f(z)^{n-1}+\dots+(z+1)^{n-1}a_1(z^3)f(z)^1+(z+1)^na_0(z^3)=0.$$$$

By irreducibility, the coefficients of our first and last equations for $$f(z)$$ must coincide. Therefore, we have $$a_i(z)=(z+1)^{n-i}a_i(z^3)$$ for $$i=0,1,\dots,n-1$$. For such an $$i$$, write $$a_i(z)=c(z)/d(z)$$, where $$c(z),d(z)\in\mathbb C[z]$$, $$d(z)$$ is not the zero polynomial, and $$c(z)$$ and $$d(z)$$ are relatively prime. Then it follows that $$c(z)d(z^3)=(z+1)^{n-i}c(z^3)d(z).$$ Suppose that $$c(z)$$ is not the zero polynomial. For all polynomials $$q\in\mathbb C[z]$$ which are not the zero polynomial, the multiplicity of the root $$-1$$ of $$q(z)$$ equals the multiplicity of the zero $$-1$$ of $$q(z^3)$$. Applying this to $$c(z)$$ and $$d(z)$$, we see that the multiplicity of the root $$-1$$ is the same in $$c(z)d(z^3)$$ as in $$c(z^3)d(z)$$. But this is a contradiction since $$(z+1)^{n-i}$$ also has at least one root $$-1$$. This proves that $$c(z)$$ is the zero polynomial, hence $$a_i(z)$$ is the zero polynomial.

Since $$a_i(z)=0$$ for $$i=0,\dots,n-1$$, we see that $$(f(z))^n+0+\dots+0+0=0$$. Therefore, $$f(z)=0$$, which is a contradiction. Therefore, $$f(z)$$ is transcendental over $$\mathbb C(z)$$. As a corollary, $$f(z)$$ is a transcendental function.

We have proved that $$f(z)$$ fulfills all the requirements, so we can apply the theorem on $$f(z)$$. Take $$\alpha:=\frac12$$. Then for all $$k\in\mathbb Z_{\geq0}$$, we have $$a(\alpha^{p^k})=1+(\frac12)^{3^k}\neq0$$. Therefore, $$f(\frac12)$$ is transcendental, and $$f\left(\frac12\right)=\prod_{i=0}^{\infty}\left(1+\left(\frac12\right)^{3^i}\right)=\prod_{i=0}^{\infty}\left(1+\frac{1}{2^{3^i}}\right).$$