Is $\sum_{k=1}^{n} \cos(k^ {1/3} )$ bounded by a constant M? I understand what $\sum_{k=1}^{n} \cos(k)$ is bounded by a constant, but I don't have any idea how to prove if $\sum_{k=1}^{n} \cos(k^ {1/3} )$ is bounded or not.
 A: Not bounded. The blocks where $\cos(^{1/3})>1/2$
 have length going to $\infty$.

First, $\cos(z)>1/2$ if $-\pi/3 < z < \pi/3$.
For each $n \in \mathbb Z$ we have $\cos(z)>1/2$ if $2\pi n-\pi/3 < z < 2\pi n+\pi/3$.
$\cos(k^{1/3}) > 1/2$ if $2\pi n-\pi/3 < k^{1/3} < 2\pi n+\pi/3$
$\cos(k^{1/3}) > 1/2$ if $(2\pi n-\pi/3)^3 < k < (2\pi n+\pi/3)^3$
The interval from $(2\pi n-\pi/3)^3$ to $(2\pi n+\pi/3)^3$ has length
$$
\left(2\pi n+\frac{\pi}{3}\right)^3 - \left(2\pi n-\frac{\pi}{3}\right)^3 = 
8\pi^3 n^2 + \frac{2\pi^3}{27}
$$
The number of integers in the inverval from $(2\pi n-\pi/3)^3$ to $(2\pi n+\pi/3)^3$ is at least
$$
8\pi^3 n^2 + \frac{2\pi^3}{27} - 2.
$$
So: consider the difference between $\sum_{k=1}^{N_1}\cos(k^{1/3})$
and $\sum_{k=1}^{N_2}\cos(k^{1/3})$ with these two values of $N$:
$$
N_1 = \left\lfloor\left(2\pi n-\frac{\pi}{3} \right)^3\right\rfloor\quad
\text{and}\quad
N_2 = \left\lfloor\left(2\pi n+\frac{\pi}{3}\right)^3\right\rfloor
$$
The difference will be
$$
\sum_{1}^{N_2} \cos(k^{1/3}) - \sum_{1}^{N_1} \cos(k^{1/3}) =
\sum_{k=N_1+1}^{N_2} \cos(k^{1/3}) > \sum_{k=N_1+1}^{N_2} \frac{1}{2} \ge 4\pi^3 n^2+\frac{\pi^3}{27}-1
$$
And this goes to $\infty$.  Therefore,
$$
\text{either}\quad
\limsup_{N \to \infty}\sum_{1}^{N} \cos(k^{1/3}) = +\infty
\quad\text{or}\quad
\liminf_{N \to \infty}\sum_{1}^{N} \cos(k^{1/3}) = -\infty
$$
or, likely, both.
