11
$\begingroup$

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?

$\endgroup$
2
  • 3
    $\begingroup$ Rational independence is not enough, but the ratio being a non-Liouvillian number is because then you get a power gain over the trivial bound for the partial sums (the same trick as in the equidistribution story, just with explicit bounds). Since $\pi$ is known to be non-Liouvillian, the answer to the original question is yes. $\endgroup$
    – fedja
    Aug 24, 2016 at 21:36
  • $\begingroup$ Sounds nice. Thanks, I'll check it out. $\endgroup$
    – H. H. Rugh
    Aug 24, 2016 at 21:47

1 Answer 1

11
$\begingroup$

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 $C_{3}( n^{2}/q+n\ln q) \leq C_{3} (n^{2-a}+n\ln n)$.

$\endgroup$
2
  • 3
    $\begingroup$ I believe this trick is known as the Van der Corput trick. It forms the basis of many bounds on exponential series, and is used extensively in additive combinatorics, subsequence ergodic theorems etc. $\endgroup$ Aug 24, 2016 at 22:58
  • $\begingroup$ This answer does not handle the case $x\in\Bbb Q$ (otherwise $\|2hx\|$ may well be zero). $\endgroup$
    – FShrike
    Jan 14 at 11:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy