Let $0<\alpha<1$. Prove that $\exists \, 0Let $[x]$ denote the fractional part of x. I'm quite lost about how to solve this problem. I suspect the solution is elementary, but all I can determine is that $x\notin\Bbb{Q}$.
 A: Let any $\alpha$ be given in the interval $(0, 1),$ or indeed in $[0,1).$
For any $x$ in $[0, 1),$ and any positive integer $n,$ we say that $x$ fails for $n$ if
$$
[nx] \leqslant \alpha^n.
$$
For any $x$ in $[0, 1),$ we say simply that $x$ fails if $x$ fails for some $n.$
For each value of $n,$ the set of "failing" values of $x$ is a union of $n$ closed, pairwise disjoint subintervals of $[0, 1),$ each of length $\alpha^n/n$:
$$
\bigcup_{i=1}^n \left[ \left(1 - \frac{i}{n}\right),
\left(1 - \frac{i}{n} + \frac{\alpha^n}{n} \right)\right].
$$
Choose a positive integer $N$ large enough to ensure that
$$
2\alpha^N < 1 - \alpha,
$$
and consider the subinterval of $(0, 1)$
$$
I = \left( \left( 1 - \frac{1}{N} \right), 1 \right),
$$
and the intersections with $I$ of the sets of failing values (of $x$) for all $n \geqslant N.$
For given $n \geqslant N,$ let there be $k$ subintervals of failing values having non-empty intersections with the open interval $I.$
If $k \geqslant 2,$ then the lower endpoint of the second-smallest of these subintervals is $1 - (k - 1)/n,$ so we must have:
\begin{gather*}
1 - \frac{k - 1}{n} > 1 - \frac{1}{N}, \\
\text{i.e.} \quad k < 1 + \frac{n}{N}.
\end{gather*}
This inequality certainly also holds if $k = 1.$
Therefore, the total length of all of the (finitely many) intervals of failing values (of $x$) for $n$ in the subinterval $I$ of $(0, 1)$ is at most:
$$
k\frac{\alpha^n}{n} \leqslant
\left( 1 + \frac{n}{N} \right) \frac{\alpha^n}{n} =
\left( \frac{1}{n} + \frac{1}{N} \right) \alpha^n \leqslant
\frac{2}{N} \alpha^n.
$$
Summing over all $n \geqslant N,$ therefore, the measure of the set of all failing values of $x$ in the subinterval $I$ is at most:
$$
\frac{2}{N} \sum_{n=N}^\infty \alpha^n =
\frac{2}{N} \frac{\alpha^N}{1 - \alpha} <
\frac{1}{N}.
$$
Because $I$ has length $1/N,$ it follows that $I,$ and therefore $(0, 1),$ contains some values of $x$ which do not fail for any $n.$
Q.E.D.
A: Have you tried proof by contradiction? Suppose there is a positive $\alpha$ less than 1 such that for all positive $x$ less than 1 there exists an $n\in\mathbb{N}$ such that $\alpha^n\ge[nx]$. Perhaps for $x$ some function of $\alpha$ this leads to a contradiction? Such as perhaps $x=1-\alpha$ although that is just a random suggestion.
