# proof of a bounded sequence satisfying a certain property

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 $$k-l$$.

• Solution on AoPS: artofproblemsolving.com/community/c7h489838p2746184 Feb 14, 2022 at 19:26
• 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. Feb 15, 2022 at 12:12

## 1 Answer

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.