What type of number is x? Problem. Let $\left \{ a_{n} \right \}_{n\in \mathbb{N}}$ be a sequence where $a_{n}\in \mathbb{N}$, $a_{n}<a_{n+1}$, $a_{n+1}\leqslant 10\cdot a_{n}$ for $\forall n\in \mathbb{N}$. What type of number is $x=0.a_{1}a_{2}a_{3}...a_{n}...$ (concatenation)? (I.e. rational or irrational)
Example: $a_{1}=1, a_{2}=10, a_{3}=100, ... a_{n}=10^{n-1}$ and $x=0.110100...10000...$
 A: I assume $x=0.a_1 a_2 a_3...a_n...$ just means to concatenate all the $\{a_{n}\}$ to form the decimal expansion of $x$.  So the number could be $$0.123456789101112...\qquad(a_{n+1}=a_{n}+1),$$ or $$0.110100100010000...\qquad(a_{n+1}=10a_{n}),$$or whatever.  What it can't be is ultimately periodic.
Let $l_{i}$ be the length (in digits) of $a_{i}$.  Since $a_{n+1}>a_{n}$, we have $a_{n}\rightarrow\infty$ and hence $l_{n}\rightarrow\infty$; and since $a_{n+1}\le 10a_{n}$, we have $l_{n+1}\le l_{n}+1$.  In other words, the sequence of lengths $\{l_n\}$ grows indefinitely without skipping any values. Now suppose $x$ is periodic with period $P$ after the $N$-th digit.  Then there must be some $M\ge N+1$ for which $l_{M}$ is a multiple of (say, $k$ times) the periodicity $P$.  Hence, $a_M$ falls in the periodic portion of $x$ (since it starts at digit $(\sum_{i=1}^{M-1}l_i) \ge M-1 \ge N$ after the decimal point), and it covers exactly $k$ repeats of the periodic substring.  Now, the next $l_{M+1}$ digits of $x$, either $kP$ or $kP+1$, must coincide with $a_{M+1}$.  But neither of these cases is possible: the next $kP$ digits are just $a_{M}$ again, and we know that $a_{M+1}>a_{M}$; while the next $kP+1$ digits are $a_{M}$ with a single nonzero digit (the first digit of $a_{M}$) appended, i.e., a number that is strictly greater than $10a_{M}$, and we know that $a_{M+1}\le 10a_{M}$.
We've shown that $x$ can't be ultimately periodic, and hence it must be irrational.
