# Is $\sum_{n\ge0}\frac1{2^{2^n}}$ rational? [duplicate]

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.

## marked as duplicate by user21820, Claude Leibovici, Did real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 25 '17 at 12:52

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.