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Let us suppose that $\{\alpha_{n}\}_{n \in \mathbb{N}}$ is a strictly increasing sequence of natural numbers and that the number obtained by concatenating the decimal representations of the elements of $\{\alpha_{n}\}_{n \in \mathbb{N}}$ after the decimal point, i.e.,

$0.\alpha_{1}\alpha_{2}\alpha_{3}\ldots$

has period $s$ (e.g., $0.12 \mathbf{12} \mathrm{121212}...$ has period 2).

If $a_{k}$ denotes the number of elements in $\{\alpha_{n}\}_{n \in \mathbb{N}}$ with exactly $k$ digits in their decimal representation, does the inequality

$a_{k} \leq s$

always hold?

What would be, in your opinion, the right way to approach this question? I've tried a proof by exhaustion without much success. I'd really appreciate any (self-contained) hints you can provide me with.

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    From your question it is not fully clear whether you know if the answer is yes, or no, or if you want to prove that $\alpha_k \leq s$ actually holds. If you don't know, the best idea would seem to be to look for a counterexample. Also, it may be more accurate to say that the concatenated sequence has _length_ $\leq s $2011-07-27
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    Also, if understood the problem well, it would seem you can get a counterexample by using s=1, and fiding sequences that generate irrational numbers; I think concatenating the elements of an arithmetic progression would do.2011-07-27
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    @gary: I think you misunderstood the question; see the two answers given.2011-07-27

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