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Determine with proof whether there exists a bounded sequence $\{a_n\}_{n\ge 1}$ of real numbers such that for all $k > l \ge 1$ we have $|a_k-a_l|\ge \frac{1}{k-l}$.

I think the statement is true. Consider the sequence defined by $a_0 = 0, a_1 = 2$ and for $0\leq k < 2^n, a_{2^n+k} = a_k+\frac{1}{2^{n-1}}.$ I tried proving the claim by induction on $k - l$, with the base case being $k-l=1$. To show this, perhaps it might be useful to assume for a contradiction that there are two consecutive terms whose absolute difference is less than $1$? Then maybe split the proof up into cases depending on whether $0\leq k,k+1 < 2^n$ or $k = 2^n-1$ for some $n\ge 1$. Assuming the base case holds, however, let $1\leq k < l$ and find $n_1$ and $m_1$ so that $2^{n_1} \leq l < 2^{n_1+1}, 2^{m_1} \leq k < 2^{m_1+1}.$ Then $a_l = a_{l-2^{n_1}} + \frac{1}{2^{n_1-1}}$. But the problem is it's not clear whether the difference between $l-2^{n_1}$ and $k-2^{m_1}$ is smaller than $k-l$.

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  • $\begingroup$ Solution on AoPS: artofproblemsolving.com/community/c7h489838p2746184 $\endgroup$
    – Martin R
    Feb 14, 2022 at 19:26
  • $\begingroup$ I did not know about the solution on AoPS when I found and wrote mine. In turn, the solution on AoPS does not mention that approximation properties of square roots and, more general, algebraic numbers are classical. $\endgroup$
    – Helmut
    Feb 15, 2022 at 12:12

1 Answer 1

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The construction of the question does not work. We have $a_{2^n+1}=2+1/2^{n-1}$ and, by induction, $a_{2^n-2}=2-1/2^{n-2}$ for $n\geq2$. Hence $a_{2^n+1}-a_{2^n-2}=3/2^{n-1}$ is not larger than $1/3$ for $n\geq5$. Multiplication of the whole sequence by a constant does not help either.

Such a sequence exists, however. It is well known that for all positive integers $p,q$ we have $$\tag1\left|\sqrt2-\frac pq\right|\geq \frac1{3q^2}.$$ This suggests to put $$b_n=\{n\sqrt2\}\mbox{ for positive integers }n,$$ where $\{x\}$ denotes the fractional part of $x$. Then $n\sqrt2-b_n=:m_m$ is an integer for all $n$. Hence by (1) $$|b_n-b_k|=|(n-k)\sqrt2-m_n+m_k|\geq\frac1{3|n-k|}.$$ Therefore the sequence $a_n=3b_n$ satisfies the wanted condition.

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