If the answer is yes, can one replace log n by another sequence that approaches ∞ faster than log n ?
Good search, Romeo! The examples provided satisfy the unboundedness condition in our question. My examples also satisfy that condition.
Thank you for the increased interest in this question, I'll leave it open of course, time is not of the essence here...
Divide [1, 2] in 10 equal length intervals and place x(1) to x(10) in each. Divide
[2, 3] in 100 equal length intervals and place x(11) to x(110) in each and so on.
The x(n) / log n goes to 0.
To George Stoica:
It seems that your question is very intriguing. It is known that the sequence {log n} (n = 1, 2, …) is not uniformly distributed modulo 1 (see, e.g., Example 2.4 in https://math.rice.edu//~michael/teaching/426_Spr14/UDmod1A.pdf). On the other hand, by my knowledge, there is a Problem which asserts that there exists a rearrangement {log (f(n))} (n = 1, 2, …) (with a bijection f: {1, 2, ….} → {1, 2, ….}) of the sequence {log n} (n = 1, 2, …) which is uniformly distributed modulo 1. If f(n) = O(log n), then the sequence {f(n)/log n} (n = 1, 2, …) is bounded.
Fejér's Theorem (see, e.g., Corollary 2.1 on page 14 of the book: L. Kuipers and H. Niederreiter, Uniform Distribution of Sequences, John Wiley & Sons, 1974). Let f(x) be a function defined for x ≥ 1 that is differentiable for x ≥ x_0. If f '(x) tends monotonically to 0 as x → ∞ and if lim_(x → ∞) x| f'(x)| = ∞, then the sequence (f(n)), n = 1, 2,...is uniformly distributed modulo 1.
As an immediate consequence of Fejér's Theorem, we have the following result (see Example 2.7 on pages 14-15 of the above book).
The following types of sequences are uniformly distributed modulo 1:
(i) (anblogt n), n = 2, 3,..., with a ≠ 0, 0 < b < 1, and arbitrary t;
(ii) (a logt n), n = 1, 2,..., with a ≠ 0 and t > 1;
(iii) (an logt n), n = 2, 3,..., with a ≠ 0 and t < 0.
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