# Is $\frac{1}{2^{2^{0}}}+\frac{1}{2^{2^{1}}}+\frac{1}{2^{2^{2}}}+\frac{1}{2^{2^{3}}}+…$ irrational?

$$\frac{1}{2^{2^{0}}}+\frac{1}{2^{2^{1}}}+\frac{1}{2^{2^{2}}}+\frac{1}{2^{2^{3}}}+\cdots$$

Is this infinite sum irrational? Is there a known way to prove it?

• A useful keyword "lacunary series" – Gottfried Helms Dec 17 '15 at 22:02

$\newcommand{\abs}{\left\lvert{#1}\right\rvert}$ A slightly different way to prove its irrationality could rely on this trick (basically the case $n=1$ of Liouville's theorem on diophantine approximation, just much less powerful):

Observation: Let $\xi\in\Bbb Q$. Let $\xi=\frac pq$, with $q>0$ and $p,q$ coprime integers. Let $m,n$ integers such that $n>0$ and $\frac{m}{n}\neq\xi$. Then $$\abs{\xi-\frac mn}\ge\frac1{qn}$$

Proof: Indeed, \begin{align}\abs{\dfrac pq-\dfrac mn}=\dfrac{\abs{np-qm}}{qn}\ge\frac1{qn}&&\text{since }np-qm\in\Bbb Z\setminus\{0\}\end{align}

Corollary: If $\xi\in\mathbb Q$, there exists a constant $\beta>0$ depending only on $\xi$ such that, whenever $\frac{m}{n}\neq\xi$, it holds $\abs{\xi-\dfrac mn}\ge\dfrac\beta n$

Now, back to your case: let $\xi:=\sum\limits_{k=0}^\infty 2^{-2^k}$ and let $\xi_n:=\sum\limits_{k=0}^n 2^{-2^k}$

You can see that $$\xi_n=\frac{\sum_{k=0}^n 2^{2^n-2^k}}{2^{2^n}}\\ 0<\abs{\xi-\xi_n}=\sum_{k=n+1}^\infty 2^{-2^k}\le\sum_{h=2^{n+1}}^\infty2^{-h}= \frac{2}{2^{2^{n+1}}}=\frac{2}{\left(2^{2^n}\right)^2}$$

Since $\dfrac{\beta}{2^{2^n}}\le\abs{\xi-\xi_n}\le\dfrac{2}{\left(2^{2^n}\right)^2}$ cannot hold definitely whatever the positive constant $\beta$, the corollary above yields that $\xi$ cannot be rational.

Yes, it is irrational. Its binary "decimal" expansion is $0.1101000100000001\ldots$. It doesn't matter what base you use, a rational number always has a periodic expansion, and this expansion clearly isn't.

• I probably need my morning coffee. But is that 'clearly' true? – Simon S Dec 17 '15 at 21:32
• @simon: Suppose it is periodic and let the period be $p$. Then since the binary expansion has more than $p$ zeros in a row (after the contribution of the $\frac1{2^{2^p}}$ term, for example) then it must be all zeroes, an absurdity. – MJD Dec 17 '15 at 22:02
• @SimonS, the strings of $0$'s between the $1$'s get long and longer, without bound. That doesn't happen in a periodic expansion. (Or what MJD says.) – Barry Cipra Dec 17 '15 at 22:03
• @MJD of course, thanks – Simon S Dec 17 '15 at 22:16