Prove That $\displaystyle\sum_{n=1}^\infty \sin(n^p)$Diverges For All $p>0$ Prove that the series $\displaystyle\sum_{n=1}^\infty \sin(n^p)$diverges for all $p>0$.
This should be simple but I have been failing... My latest attempt is Cauchy's criterion.
 A: I don't know the answer to the question, and I think it is a mistake in the question itself. Here are a few remarks.
So, when is $\sum_{n \geq 0} \sin (n^p)$ convergent?


*

*If $p < -1$: the series is absolutely convergent.


This is because $\sin (n^p) \sim n^p$, which is the main term of an absolutely convergent series.


*

*If $p \in [-1,0]$: the series diverges to $+ \infty$.


If $p \in [-1,0)$, we have $\sin (n^p) \sim n^p$, which is positive and the main term of a divergent series. If $p = 0$, the partial sums are $n \sin (1)$, which goes to infinity.


*

*If $p \in (0,1)$: the series is divergent.


We have $(n+1)^p - n^p \sim p n^{p-1}$. If $n^p \equiv x$, then $(n+1)^p - n^p \equiv p x^{1-\frac{1}{p}}$. In particular, since the steps are this small, for some constant $C$, for all $x$, you will find an $n$ such that $n^p$ is at distance at most $C p x^{1-\frac{1}{p}}$ of $x$. Next, take $x_k := \pi/2+2\pi k$; this gets you an increasing sequence of integers $n_k$ such that $\lim_{k \to + \infty} \sin (n_k^p) = 1$.
Since the sequence $(\sin (n^p))_{n \geq 0}$ does not converges to $0$, the series is divergent. My best guess is that the partial sums will oscillate more and more, so that the partial sums do not diverge to $+ \infty$ or $- \infty$.


*

*If $p \geq 1$ is rational: the series is divergent.


It is enough to prove that $\sin (n^p)$ does not converges to $0$. For any $n \geq 0$, write:
$$n^p = 2 \pi k_n + \varepsilon_n,$$
where $k_n$ is an integer and $\varepsilon_n \in [-\pi, \pi)$. Since $p$ is rational, there exists and integer $\lambda \geq 2$ such that $\lambda^p$ is an integer. Note that:
$$(\lambda^m n)^p = 2 \pi k_n \lambda^{mp} + \lambda^{mp} \varepsilon_n.$$
Since $\pi$ is transcendental, $\varepsilon_n \neq 0$ for all $n > 0$. Hence, 
I can find an integer $m$ such that $|\lambda^{mp} \varepsilon_n| \in [\lambda^{-p}\pi,\pi]$, so that:
$$|\sin ((\lambda^m n)^p)| \geq \sin (\lambda^{-p}\pi).$$
Hence, for any $n>0$, there exists $n' \geq n$ such that $|\sin ((n')^p)| \geq \sin (\lambda^{-p}\pi)$. I can then construct recursively an increasing sequence of integers $(n_k)_{k \geq 0}$ such that $|\sin (n_k^p)| \geq \sin (\lambda^{-p}\pi)$ for all $k$, and the sequence $(\sin (n^p))_{n \geq 0}$ does not converge to $0$.


*

*In general: I have no idea. There may be strange phenomena if we tune $p$ (see e.g. Mill's constant for a related construction).

A: I think it is simple to see that the limit $\lim_{\infty}\sin{n^p}$ does mot go to $0$ (does not even exists) therefore the series 
$$ \sum_{n=0}^{\infty}\sin{n^p}$$ diverges.
Recall
For every converging series $ \sum_{n=0}^{\infty}a_{n}$ we have $a_n\to 0,~~$ as $n\to \infty$
