Prove that $\sum_{k=0}^\infty \cos(k^2t)$ diverges for all $t \geq 0$ I think the series does not converge because the sequence $\cos(k^2t)$ doesn't converge to $0$. However, I cannot prove this last fact. I have tried looking for a subsequence of $k^2t$ such that for each $k$ there is an integer $z$ such that $z\pi-\frac{\pi}{4} \leq k^2t \leq z\pi+\frac{\pi}{4}$, but I haven't been able to show such a sequence exists. I have also tried some approaches using Chebyshev's formula, points in $\mathbb{R}^2$, and Taylor's expansion for $\cos(x)$, all unsuccessfuly.
 A: If $\cos (k^2 t)$ is small, then $k^2 t$ is close to an odd multiple of $\frac \pi2$. But then, $(2k)^2 t$ is close to an even multiple of $\frac\pi2$, and $\cos((2k)^2 t)$ will be large.
So there will never be an integer $N$ such that, for all $k > N$, $|\cos(k^2t)|$ is less than, say, $0.1$. Whenever $\cos(k^2t)$ achieves a value that small, $\cos((2k)^2 t)$ will be correspondingly large: at least $\cos(4 \arcsin 0.1) =0.9208$.
A: It is convenient to consider $2 \pi t$ in place of $t$:
$$ \sum_{k=0}^{\infty} \cos(2 \pi k^2 t). $$


*

*If $t \in \Bbb{Q}$, then the summand is not identically zero and is periodic in $k$. So $\cos(2 \pi k^2 t)$ does not converge to $0$ and hence the series diverges.

*If $t \notin \Bbb{Q}$, then it is well-known that $k \mapsto k^2 t \text{ mod } 1$ is equidistributed on $[0, 1]$. Again, $\cos(2 \pi k^2 t)$ does not converge to $0$ and hence the series diverges.
A: $A_T:=\{ s|s=k^2t\ {\rm mod}\ 2\pi$ and $k\leq T \}$. If $T$ is larger, then $A_T$ will be almost uniformly distributed in $[0,2\pi)$ : $$ \lim_T\ \frac{\sharp \ [a,b]\cap A_T }{\sharp \ [A,B]\cap A_T} =\frac{b-a}{B-A}$$ for all $[a,b],\ [A,B]\subset [0,2\pi)$
