Non-repeating decimal expansion of known number

Question

The decimal expansion of any irrational number $x>0$ is non-repeating. This is well known. So, we have a way of obtaining irrational numbers, such has$$x=0.101\,101\,110\,111\,101\,111\,101\ldots$$(after the decimal point, we have one $1$, one $0$, two $1$'s, one $0$, three $1$'s, one $0$, and so on).

My (admittedly vague) question is this: how to obtain a known irrational number whose decimal expansion (or, for that matter, whose expansion on some base) is easily shown to be non-repeating? Of course, when I write that that the decimal expansion “is easily shown to be non-repeating” what I mean is that it is easy to describe its decimal expansion (and to see that it is non-repeating); otherwise, one could just say that, since the number is irrational, its decimal expansion must be non-repeating. And by “known number” I mean something like, say, $\sqrt[3]2$ or $\pi^e$.

Answer 1

The example that you posted $$x=0.10110111011110...$$ Can be expressed as an infinite series. As you can see, the position of the $n$th zero in this decimal representation is given by $$\frac{(n+1)(n+2)}{2}-1$$ and so the value of your number is equal to $$x=0.\overline1 -\sum_{n=1}^\infty 10^{1-\frac{(n+1)(n+2)}{2}}$$ Since $0.\overline1=\frac{1}{9}$, then we have $$x=\frac{1}{9}-\sum_{n=1}^\infty 10^{1-\frac{(n+1)(n+2)}{2}}$$ Which can be simplified to $$x=\frac{1}{9}-\sum_{n=1}^\infty (\sqrt{10})^{-n^2-3n}$$ As far as I know, this is as much as it can be simplified without bringing in other non-elementary functions. However, you might be able to use this function to simplify it further.