Conditional convergence of $\sum_{n\geq 1} \frac{\sin(p(n))}{n}$?

Question

The series $\sum_{n\geq 1} \frac{\sin n}{n}$ is easily seen to be conditionally convergent, e.g. by Abel summation. But how about $\sum_{n\geq 1} \frac{\sin(n^2)}{n}$? (for which Abel summation fails)

I came across the problem of showing that $\sum_{n\geq 1} \frac{|\sin (n^2)|}{n}$ is divergent in convergence of $\sum_{n\geq 1}|\sin(n^2)|/{n}$. A natural question is if it converges conditionally without the absolute signs.

More generally, if given a polynomial $p(n)$ where the dominant coefficient is rationally independent from $\pi$ then by Weyl, see e.g. Tao-Weyl equidistribution $p(n)$ is equidistributed (mod 1) but is the equidistribution sufficiently rapid for the series $\sum_{n\geq 1} \sin(p(n))/n$ to be conditionally convergent?

Answer 1

This is something I learned from fedja (artofproblemsolving.com):

We will show that if $x$ is not Liouvillian then the sum $$ \sum_{n \geq 1} \frac{\sin (2 \pi x n^{2})}{n} $$ converges. An argument goes like this: rewrite $$ \sum_{k =1}^{n} \frac{\sin (2 \pi x k^{2})}{k} = \frac{1}{n}\sum_{k=1}^{n}\sin (2 \pi x k^{2}) + \sum_{k=1}^{n} \frac{S_{k}}{k(k+1)}=I+II $$ where $S_{k} = \sum_{j=1}^{k}\sin (2 \pi x k^{2})$. The first term is OK (converges to integral). For the second term it is enough to show that $|S_{k}| = O(k^{\delta})$ for some $0<\delta<1$. We use the following trick:

$$ S_{n}^{2} \leq \left| \sum_{k=0}^{n} e^{2 \pi i x k^{2}}\right|^{2} = \sum_{k,m} e^{2\pi i x(k-m)(k+m)} = (*) $$ Next change variables $k-m=h$ and $m=m$ then

$$ (*) =\sum_{-n \leq h \leq n}e^{2 \pi i x h^{2}} \sum_{1\leq m \leq n, 1 \leq m+h \leq n}e^{2 \pi i \cdot 2hxm} \leq \sum_{-n \leq h \leq n}\left| \sum_{1\leq m \leq n, 1 \leq m+h \leq n}e^{2 \pi i \cdot 2hxm}\right| $$ Clearly (Dirichlet kernel)

$$ \left| \sum_{1\leq m \leq n, 1 \leq m+h \leq n}e^{2 \pi im \cdot 2hx}\right| \leq \frac{C}{\|2hx\|} $$ where $\|2hx\| = |2hx \mod 1|$ is the distance to the nearest integer. Since $x$ is not Liouvillian then there exists $0<q\leq n$ such that $|2x - \frac{p}{q}| \leq \frac{1}{nq}$ and $q>n^{a}$ for some $a>0$. This means that when $h \in [-n, n]$ then $2hx$ is close to the fraction of the type $h p /q$ up to an error $|h/qn|\leq 1/n^{a}$ therefore when $h$ runs over $[-n, n]$ then $2hx \mod 1$ almost runs over the fractions $\ell/q$, $\ell <q/2$ and each of them may appear at most $10 n/q$ times. Therefore $$ \sum_{-n \leq h \leq n} \frac{C}{\|2hx\|}\leq C_{2} \frac{n}{q}\left( n+\sum_{\ell=1}^{q/2}\frac{q}{\ell}\right) $$

(The first $n$ in the parenthesis comes from the error). And everything is bounded as $C_{3}( n^{2}/q+n\ln q) \leq C_{3} (n^{2-a}+n\ln n)$.