Is $\sum_{k=1}^{n} \cos(k^ {1/3} )$ bounded by a constant M?

Question

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.

Answer 1

Not bounded. The blocks where $\cos(^{1/3})>1/2$ have length going to $\infty$.

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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.