TITLE: $0.101001000100001000001$... is irrational.
QUESTION [15 upvotes]: How do I show that $0.101001000100001000001...$ is irrational? How do I generalize this to decimal expansions of a similar type, i.e. what is a criterion for decimal expansions with $0$'s and $1$'s involving long gaps of zeros and which guarantees irrationality?

REPLY [2 votes]: The given number $\displaystyle\;x = \sum_{n=1}^\infty 10^{-n(n+1)/2}\;$ is irrational.
Given any rational number $r > 0$, we can rewrite it into the form
$$r = \frac{1}{10^a}\left(b + \frac{p}{q}\right)$$
 where $a, b, p, q \in \mathbb{N}$, $0 \le p < q$ and $\gcd(10,q) = 1$. If $p =0$ or $q = 1$, everything is obvious.
Consider the case $0 < p < q$ and let $\varphi(q)$ be the
Euler's totient function.
By Fermat-Euler theorem, we have
$$10^{\varphi(q)} \equiv 1 \pmod q 
\quad\implies\quad
10^{\varphi(q)} - 1 = qs \quad\text{ for some integer } s$$
This leads to a decimal expansion of $\displaystyle\;\frac{1}{q}$, 
$$\frac{1}{q} = \frac{s}{10^{\varphi(q)} - 1} = \sum_{k=1}^\infty s \times 10^{-k\varphi(q)}$$
which repeats every $\varphi(q)$ digits.
Multiply this expansion by $p$ gives us a decimal expansion of $\frac{p}{q}$ which again repeat every $\varphi(q)$ digits. Adding $b$ and then divide by
$10^a$ will introduce some garbage in the front and shift the whole expansion
to the right for $a$ digits. At the end, we continue to have an expansion which is
ultimately periodic with a period which is a divisor of $\varphi(q)$.
Back to our number $x$, one find the sequence of '$0$' in its decimal expansion can be arbitrary long. Given any $q$, one can always a find a sequence of '$0$' longer than $\varphi(q)$. This rules out the possibilities that $x$ is 
a rational number with a corresponding $q$. As a result, $x$ cannot be rational.
In fact, the number $x$ is known to be transcendental. See the refs in my answer to a related question.