More on Sequences — Question 3

*Question 3 Let $a_n=\{n\sqrt2\}=n\sqrt2-\lfloor n\sqrt2\rfloor$.

1. Prove that $0\le a_n<1$, so the sequence is bounded.

2. Using the fact that the fractional parts of multiples of an irrational number are dense in $[0,1]$, find two different subsequential limits.

3. Decide whether $(a_n)$ converges and explain why boundedness is insufficient.

*Question 3 – Solution

For every real $x$, its fractional part satisfies $0\le\{x\}<1$. Therefore $0\le a_n<1$ for all $n$, and the sequence is bounded.

Because $\sqrt2$ is irrational, the density theorem for irrational rotations says that the set $\{\{n\sqrt2\}:n\ge1\}$ is dense in $[0,1]$. In particular, every open subinterval contains infinitely many terms, allowing the selection of increasing subsequence indices.

For each positive integer $j$, choose increasing indices $n_j$ and $m_j$ such that $$0<a_{n_j}<1/j,\qquad 1/2-1/j<a_{m_j}<1/2+1/j.$$ The first inequality and Squeeze Theorem give $a_{n_j}\to0$; the second gives $a_{m_j}\to1/2$.

Every subsequence of a convergent sequence must approach the same limit as the original sequence. Since these two subsequences have different limits, $(a_n)$ diverges. Boundedness alone guarantees neither monotonicity nor convergence.