If $\lim_{x\to \infty} f^{(k)}(x)=\theta$ (irrational), then $(f(n))$ is uniformly distributed modulo $1$ 
Let $f\in \mathcal C^{k+1}$, $x\ge 1$ and let for some integer $k\ge 1$, we have
$$\lim_{x\to \infty} f^{(k)}(x)=\theta$$
for irrational $\theta$. Then, prove that $(f(n))$ is uniformly distributed modulo $1$.

To do this, I think, I need to prove the following proposition first-

Let $(x_n)$ be a sequence of real numbers with the property that for some integer $k\ge 1$,
$$\lim_{n\to \infty} \Delta^k x_n=\theta$$
where $\theta$ is irrational. Then, prove that $(x_n)$ is uniformly distributed modulo $1$.

I do know that if
$$\lim_{n\to \infty} \Delta x_n=x_n-x_{n-1}=\theta$$
for irrational $\theta$, then $(x_n)$ is ud mod $1$. I have also used this (and IVT) to show that if $f\in \mathcal C^2$, $x\ge 1$ and if
$$\lim_{x\to \infty} f^\prime (x)=\theta$$
for irrational $\theta$, then $(f(n))$ is ud mod $1$.
To prove the proposition, I think, I need to use the stated result and use Induction by noting that if $h$ is a positive integer, then
$$\Delta^k\left(f(n+h)-f(n)\right)=\sum_{j=0}^{h-1}\Delta^{k+1}f(n+j)$$
I feel like I may also have to use the fact that if $(x_n)$ is a sequence such that $(x_{n+h}-x_n)$ is ud mod $1$ for all positive integers $h$, then $(x_n)$ is ud mod $1$.
However, I couldn't proceed any further.
 A: Let $a_n = f(n)$. Let $e(x) = e^{2\pi i x}$. All equidistribution is modulo $1$.
This is the consequence of the following result.

Theorem: Let $\{a_n\}$ be a sequence. If $\{a_{n + h} - a_n\}_n$ is equidistributed for any fixed positive integer $h \geq 0$, then $\{a_n\}$ is equidistributed.

For the proof, complete Chapter 4, Problem 2, Fourier Analysis by Stein-Shakarchi.
For this problem, we argue by induction on $k$. We first state the induction step for $k \geq 2$: if $f(x)$ satisfies the condition of the problem, then for fixed $h \in \mathbb{N}$, the function $g_h(x) = f(x + h) - f(x)$ satisfies
$$g_h^{(k - 1)}(x) = f^{(k - 1)}(x+ h) - f^{(k - 1)}(x) = h f^{(k)}(\xi), \xi \in [x, x + h].$$
So we have
$$\lim_{x \to \infty} g_h^{(k - 1)}(x) = h\theta.$$
So $g_h$ satisfies the induction hypothesis and $\{f(n + h) - f(n)\}$ is equidistributed, thus $\{f(n)\}$ is equidistributed by the theorem.
The case $k = 1$ needs to be dealt with separately. We use this rearrangement of sums
$$\sum_{n = 1}^N e(a_n) = \sum_{k = 1}^{N / H} e(a_{kH}) \sum_{\ell = 0}^{H - 1} e(a_{kH + \ell} - a_{kH}).$$
Thus,
$$\frac{1}{N} \left\vert \sum_{n = 1}^N e(a_n)\rvert \leq \frac{1}{N/H} \sum_{k = 1}^{N / H} \frac{1}{H}\lvert \sum_{\ell = 0}^{H - 1} e(a_{kH + \ell} - a_{kH}) \right\vert .$$
For any fixed $h$, we have $a_{n + h} - a_n \to \theta$, so
$$\lim_{k \to \infty}\left\vert \sum_{\ell = 0}^{H - 1} e(a_{kH + \ell} - a_{kH}) \right\vert = \sum_{\ell = 0}^{H - 1} e(\ell \theta).$$
Thus
$$\limsup_{N \to \infty} \frac{1}{N}\left\vert \sum_{n = 1}^N e(a_n)\rvert \leq \frac{1}{H}\lvert \sum_{\ell = 0}^{H - 1} e(\ell \theta) \right\vert.$$
This holds for any $H$, so take $H \to \infty$, as $\{\ell \theta\}$ is equidistributed, we have
$$\limsup_{N \to \infty} \frac{1}{N}\left\vert \sum_{n = 1}^N e( a_n)\right\vert \leq \liminf_{H \to \infty}\frac{1}{H}\left\vert \sum_{\ell = 0}^{H - 1} e(\ell \theta) \right\vert = 0.$$
Thus we conclude that
$$\frac{1}{N}\left\vert \sum_{n = 1}^N e(a_n) \right\vert \to 0.$$
Similarly, we can show that for any positive integer $k$,
$$\frac{1}{N}\left\vert \sum_{n = 1}^N e(k a_n) \right\vert \to 0.$$
So $\{a_n\}$ is equidistributed as desired.
