Is $\sum_{n\ge0}\frac1{2^{2^n}}$ rational?

Question

How might I determine whether $\sum_{n\ge0}\frac1{2^{2^n}}$ is rational?

I have seen in the answers to this question that the sum converges and it was also mentioned that this is called Kempner's number. Another question mentions this under the name Fredholm number.

Similar: https://math.stackexchange.com/questions/2365746/prove-the-series-sum-n-1-infty-frac15n-converges-to-an-irration — Martin R, Aug 02 '17 at 12:35

score 15 · Accepted Answer · answered Aug 02 '17 at 01:21

15

In base 2, it is not finite or recurring, so it is irrational.

answered Aug 02 '17 at 01:21

Aryabhata

82,206

1

Brilliant solution. I'll definitely keep this one in my arsenal. – Cameron Williams Aug 02 '17 at 01:34

score 4 · Answer 2 · answered Aug 02 '17 at 01:38

According to this paper, if $$a_n-a_{n-1}^2+a_{n-1}-1>0$$ for all $n$ greater than some $N\in\mathbb{N}$, then $$\sum_{n=0}^{\infty}\frac{1}{a_n}$$ is irrational. In your case, $$a_n-a_{n-1}^2+a_{n-1}-1=2^{2^n}-(2^{2^{n-1}})^2+2^{2^{n-1}}-1=2^{2^{n-1}}-1$$ which is greater than $0$ for all $n\geq 0$. Thus, $$\sum_{n=0}^{\infty}\frac{1}{2^{2^n}}$$ is irrational.

Is $\sum_{n\ge0}\frac1{2^{2^n}}$ rational?

2 Answers2