An unbounded sequence s.t. $\forall t>0 |[a_nt]-a_nt|\rightarrow 0$ While studying probability I accidentally proved the result below by what I think is an overly complicated proof (a sketch of this proof below). Is there an easy way to prove it?
The result is the following.

Let $\{a_n\}_{n\in\mathbb{N}}$ be a sequence over the reals such that for all $t>0$ we have $\lim_{n} |[a_nt]-a_nt|=0$, where we denote with $[x]$ the nearest integer to x.
Then $\{a_n\}_{n\in\mathbb{N}}$ converges to $0$.

This is equivalent to another result.

Let $\{a_n\}_{n\in\mathbb{N}}$ be a sequence over the reals and $a$ a real number such that for all $t>0$ we have $\lim_{n} \text{e}^{\text{i}a_nt}=e^{\text{i}at}$
Then $\{a_n\}_{n\in\mathbb{N}}$ converges.

Proof. Let us suppose that $\{a_n\}_{n\in\mathbb{N}}$ doesn't converge. We will study two possible scenarios, both leading to contradiction.
First case. The sequence $\{a_n\}_{n\in\mathbb{N}}$ is bounded.
Then, choosing a small enough $t$, we get that $\forall n |a_nt|<\pi$, and we conclude, using the continuity of the complex logarithm, that the sequence converges to $a+2k\pi$ for some $k\in\mathbb{Z}$.
Second case. The sequence $\{a_n\}_{n\in\mathbb{N}}$ is unbounded. Then, we consider a sequence $\{X_n\}_{n\in\mathbb{N}}$ of random variables such that $\forall n X_n\sim N(a_n,1)$ and their respective characteristic functions $\{\text{exp}(\text{i}a_nt-t^2/2)\}_{n\in\mathbb{N}}$. Our initial assumptions imply that the sequence of characteristic functions is pointwise convergent to $\text{exp}(\text{i}at-t^2/2)$, so, by Paul-Levy's theorem, the sequence of random variables $\{X_n\}_{n\in\mathbb{N}}$ converges in distribution to a random variable $X$ such that $X\sim N(a,1)$. However, it can be proved via Chebishev inequality that if the sequence of numbers $\{a_n\}_{n\in\mathbb{N}}$ is unbounded then the sequence of random variables doesn't converge in distribution to any random variable.
 A: In order to redeem myself from the stupid comments the extremely late other night, I thought that I'd share a different proof of the second statement. Primarily, we notice that
$$ 2 \left(1- \frac{\sin(a_n - a)}{(a_n - a)} \right) = \int_0^1 \lvert e^{ia_nt}-e^{iat} \rvert^2 dt \rightarrow 0 \qquad,\, n\rightarrow \infty.
$$
This follows from the fact that $e^{ia_n t} \rightarrow e^{iat}$ pointwise, for all $t>0$ and the dominated convergence theorem. There might be a simpler way to prove this particular step, but for the moment we could content ourselves with this. It readily follows that
$$
\lim_{n \rightarrow \infty} \frac{\sin(a_n -a)}{(a_n - a)} = 1.
$$
For the sake of obtaining a contradiction, suppose that $a_n \nrightarrow a$, as $n \to \infty$. Then there exists a $\delta >0$ and sequence of integers $\{n_j \}$, such that $\lvert a_{n_j}-a \rvert \geq \delta$, for all $j\geq 1$. Since $t \mapsto \frac{\sin(t)}{t}$ is (uniformly) continuous with global maximum at $t=0$, there consequently exists a number $0<\eta <1$, such that
$$
\frac{\sin(a_{n_j}-a)}{(a_{n_j}-a)} \leq (1-\eta), \qquad, \, \forall j\geq 1.
$$
This implies that
$$ \bigl| 1- \frac{\sin(a_{n_j}-a)}{(a_{n_j}-a)}\bigr| = \left( 1- \frac{\sin(a_{n_j}-a)}{(a_{n_j}-a)}\right) \geq \eta \qquad, \forall j\geq 1, $$
which contradicts the previously established fact that $\frac{\sin(a_n -a)}{(a_n -a)} \to 1$, as $n\to \infty$.
A: Notation: I use $d(r, \Bbb Z)$ for the (non-negative) distance from $r\in \Bbb R$ to the nearest integer.
We show that $[1]$ if there were a counter-example, that is, if $d(ta_n, \Bbb Z)\to 0$ for all $t>0$ but $\neg (a_n\to 0)$ then there is a strictly monotonic sequence $(z_n)_n$ of positive integers such that $d(tz_n,\Bbb Z)\to 0$ for all $t>0.$
We also show that $[2]$ if $(z_n)_n$ is a strictly monotonic sequence of positive integers then there exists $t>0$ such that $d(tz_n,\Bbb z)\in [1/4,3/4]$ for infinitely many $n.$
Hence no counter-example to the main Q can exist.
Proof of $[1]$. Suppose $a=(a_n)_n$ is a counter-example. Let $b=(b_n)_n$ be a sub-sequence of $a$ with $\inf_n|b_n|>0.$
Now b cannot have any convergent sub-sequence $c=(c_n)_n$ with a limit $l. $ Otherwise, for any $t>0$ we would have $tc_n\to tl,$ but $d(tc_n,\Bbb Z)\to 0$, so $tl\in \Bbb Z,$ but $l\ne 0$ (as $|l|\ge \inf_n|b_n|>0$), so every $t\in \Bbb R^+$ would belong to the countable set $\{z/l:z\in \Bbb Z\},$ which is absurd.
So $b$ has a strictly monotonic sub-sequence $d=(d_n)_n$ diverging to $\infty$ or to $-\infty$ such that $|[d_{n+1}]|-|[d_n]|$ is strictly increasing. And $d$ has a sub-sequence $e=(e_n)_n$ such that $d(e_n,\Bbb Z)$ converges to a limit $u.$
Let $e_n=[e_n]+u+\delta_n,$ with $\delta_n\to 0.$ Finally, let $z_n=[e_{n+1}]-[e_n].$ We have $$tz_n=t([e_{n+1}]-[e_n])=$$ $$=t(\,(e_{n+1}-u-\delta_{n+1})-(e_n-u-\delta_n)\,)=$$ $$=(te_{n+1}-[te_{n+1}])-(te_n-[te_n])-t(\delta_{n+1}-\delta_n)+([te_{n+1}]-[te_n]).$$ From this, since $([te_{n+1}]-[te_n])\in\Bbb Z,$ we have $$ d(tz_n,\Bbb Z)\le   |te_{n+1}-[te_{n+1}]|+|te_n-[te_n]|+|t|\cdot |\delta_{n+1}|+|t|\cdot |\delta_n|,$$ which $\to 0$ as $n\to\infty.$
Proof of $[2]$.Let $(z_n)_n$ be a strictly increasing sequence in $\Bbb N$. We obtain $t>0$  such that $d(tz_n,\Bbb N)\in [1/4,3/4]$ for infinitely many $n$ by an approximating sequence $(t_n). $
The idea is that
(i) $\,(t_n)$ is a positive increasing sequence with limit $t$, and
(ii) $\,f:\Bbb N\to \Bbb N$ is strictly increasing, and
(iii) for each $n$ and for $1\le j\le n$ we have $1/4< t_n\cdot z_{f(j)}-v_{n,j}<3/4$ where each $v_{n,j}\in \Bbb N_0$.
If we can do this then we will have $d(tz_{f(n)},\Bbb Z)\in [1/4,3/4]$ for every $n$.
So let $f(1)=1$ and let $t_1=1/2z_1$ .  Inductively on $n$, suppose  $1/4< t_n\cdot z_{f(j)}-v_{n,j}<3/4$ when $1\le j\le n$ with each  $v_{n,j}\in \Bbb N_0.$ We find $t_{n+1}$ and $f(n+1)$  as follows:
(a) If there exists $m>f(n)$ such that $d(t_nz_m,\Bbb Z)\in (1/4,3/4)$ then let this $m=f(n+1)$ and let $t_{n+1}=t_n.$
(b) If (a) does not apply, but if there are infinitely many $m>f(n)$ such that $t_nz_m=k_{n,m}+p_{n,m}$ with $k_{n,m}\in \Bbb N_0$ and $p_{n,m}\in [0,1/4]:$ Take such an $m,$ large enough that $1/(3z_m)<2^{-n}$ and also large enough that $(t_n\cdot z_{f(j)}-v_{n,j})+1/(3z_m)<3/4$ for $1\le j\le n.$ Now let $ f(n+1)=m$ and let $t_{n+1}=t_n+1/(3z_m)$.
(c) If (a) and (b) do not apply then there are infinitely many $m>f(n)$ such that $t_nz_m=k_{n,m}+p_{n,m}$ with $k_{n,m}\in \Bbb N_0$ and $p_{n,m}\in [3/4,1).$ Take such an $m,$ large enough that $7/(12z_m)<2^{-n}$ and also large enough that $(t_n\cdot z_{f(j)}-v_{n,j})+7/(12z_m)<3/4 $ for $1\le j\le n.$ Now let $ f(n+1)=m$ and let $t_{n+1}=t_n+7/(12z_m)$.
A: Suppose $a_n$ does not converge to zero, so there is some $\varepsilon > 0$ for which infinitely many $a_n$ have $|a_n| > \varepsilon$. Let $\{a_{n_k}\}$ be the subsequence consisting of those elements with absolute value exceeding $\varepsilon$, and let $b_k = a_{n_k}/\varepsilon$, so all $|b_k| > 1$ and $|[b_k t] - b_k t| \to 0$ for all $t > 0$. Define $f(x) = |[x] - x|$, and note that $f$ is nonnegative, bounded above by $1$, periodic with period $1$, and has $\int_0^1 f(x) \,dx = \frac{1}{4}$. Then since $f(b_n x)$ is periodic with period $\frac{1}{|b_n|}$, $f(b_n x)$ achieves $\lfloor |b_n| \rfloor \geq \frac{|b_n|}{2}$ full periods on the interval $[0, 1]$, and thus
$$\int_0^1 f(b_n x) \,dx \geq \frac{\lfloor |b_n| \rfloor}{4|b_n|} \geq \frac{1}{8}.$$
Now, for each $N$, let $B_N$ be the set of $t \in (0, 1)$ for which $f(b_n t) < \frac{1}{20}$ for all $n \geq N$. Then $\{B_N\}$ is an ascending sequence of (measurable) sets, and by assumption, each $t$ is in some $B_N$, so $\bigcup_N B_N = (0, 1)$. This means $\lim_{N \to \infty} \mu(B_N) = 1$, so in particular there is some $N$ for which $B_N$ has measure at least $1 - \frac{1}{20}$. But then for this $N$ we have
$$\int_0^1 f(b_N x) \,dx \leq \int_{B_N} f(b_N x) \,dx + \int_{(0, 1) - B_N} f(b_N x) \, dx \leq \frac{1}{20} + \frac{1}{20} = \frac{1}{10}$$
contradicting our lower bound on the integral. Thus $\{a_n\}$ must converge to zero.
