4
$\begingroup$

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.

$\endgroup$

marked as duplicate by user21820, Claude Leibovici, Did real-analysis Aug 25 '17 at 12:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

15
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Brilliant solution. I'll definitely keep this one in my arsenal. $\endgroup$ – Cameron Williams Aug 2 '17 at 1:34
4
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.