I have tried to solve this type of series :
$$\sum \frac{e^{i\, u(n)}}{v(n)} $$
For some $u,v$ an Abel Transform allow to find convergence, but for $u(n)=n^2$ and $v(n)=n$ I can't find an argument.
Also if $v(n)=n$, do you think there could be a condition on $u$ to have convergence?
Not a complete answer
I'll take $v(n)=n$, since this seems to be of primary interest to you. One approach to proving convergence of the series $\sum_{n=1}^\infty \frac{1}{n} e^{i u(n)}$ is to split it into dyadic blocks $2^k\le n<2^{k+1}$ and try to get a bound of the form $$\left| \sum_{n=2^k}^{2^{k+1}-1} e^{i u(n)} \right| \lesssim 2^{(1-\epsilon) k} \tag1$$ for some $\epsilon>0$. Since $n\approx 2^k$ within this dyadic block, the series is then controlled by $\sum_k 2^{-\epsilon k}$, which converges. The hard part is to get this $\epsilon$ of improvement over the trivial bound $\le 2^{k}$.
The sum on the left of (1) is called an exponential sum. When $u$ is a polynomial, Weyl's inequality can be used. In particular, these lecture notes discuss the case of quadratic polynomial $u$. The effectiveness of Weyl's inequality for the sum $\sum e^{2\pi i\, p(n)}$ depends on the irrationality of the leading coefficient of polynomial $p$, which in your case is $1/(2\pi)$. I do not know if known results about the irrationality measure of $\pi$ will suffice for your purpose.