Prove or disprove: the sequence $a_n = \{ \alpha n \}$ (fractional part) converges if and only if $\alpha \in \mathbb{Z}$

Question

Prove or disprove: the sequence $a_n = \{ \alpha n \}$ (fractional part) converges if and only if $\alpha \in \mathbb{Z}$

I'm quite sure that this statement is true, but I'm having a little difficulty proving the forward direction. My approach is to do a proof by contradiction, assuming $\alpha \notin \mathbb{Z}$ and using the fact that if the limit is convergent, then $\lim_{n \to \infty} \{ \alpha n \} = \lim_{n \to \infty} \{ \alpha (n+1) \} = \lim_{n \to \infty} \{ \alpha n+ \alpha \}$. Now, for finite $\alpha n$ this equality would imply that $\alpha \in \mathbb{Z}$, but because $\alpha n$ is not finite I'm not sure exactly how to prove that $\alpha \in \mathbb{Z}$ here. Any advice please?

Answer 1

We may distinguish $\alpha$ according to it is rational or not.

If $\alpha=h/k$ for some coprime $h$ and $k$, then for all integer $0\le r<k$ there exists some $n_0$ such that $n_0h\equiv r\pmod k$, so setting $n_j=n_0+kj$ gives

$$ \{\alpha n_j\}=\left\{{n_0-r\over k}+j+\frac rk\right\}=\frac rk. $$

This indicates that for all $0\le r<k$, $r/k$ is a limit point of the sequence $\{\alpha n\}$. As a result, it won't converge when $k>1$. For the case $k=1$, we have trivially $\{\alpha n\}=0$.

If $\alpha$ is irrational, Kronecker's approximation theorem indicates that every point in $[0,1)$ is a limit point $\{\alpha n\}$, so it does not converge in this case.

Therefore, we conclude that $\{\alpha n\}$ converges if and only if $\alpha$ is an integer.

Answer 2

Let us write $$\alpha n=b_n+c_n$$ where $b_n=[\alpha n], c_n=\{\alpha n\}$ are integer and fractional parts of $\alpha n$. We are given that $c_n\to c$ as $n\to\infty$.

Now we have $$\alpha =\alpha(n+1)-\alpha n=(b_{n+1}-b_n)+(c_{n+1}-c_n)$$ Let us observe that $c_{n+1}-c_n\to c-c=0$ and hence $b_{n+1}-b_n\to\alpha$. Further note that $b_{n+1}-b_n$ is a sequence of integers and since it converges it must remain constant after a certain value of $n$. This constant value is the limit $\alpha $ as well as hence $\alpha$ is an integer.