Does $S_k= \sum \limits_{n=1}^{\infty}\sin(n^k)/n$ converge for all $k>0$?

Motivation: I recently learned that $S_1$ converges. I think $S_2$ converges by the integral test. Was the question known in general?

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    $\begingroup$ I am not sure you can apply the integral test here for $k=2$. Isn't the function being non-negative a necessary condition? $\endgroup$
    – Aryabhata
    Aug 12, 2010 at 19:03
  • $\begingroup$ @Moron: you might be right. I thought I could use this one: jstor.org/stable/2305031?cookieSet=1 , but may be I made a mistake. $\endgroup$
    – curious
    Aug 12, 2010 at 19:18
  • $\begingroup$ @curious I don't think so. Suppose $f(x)=\sin(x^2)/x$, we have $f'(x)=2\cos(x^2)-\sin(x^2)/x^2$. If $\int f'(x)$ converges absolutely, then $\int \cos(x^2)$ converges absolutely, contradiction! I think integral test wouldn't work if the integrand is oscillating near $\infty$. $\endgroup$
    – Yai0Phah
    May 16, 2013 at 16:43
  • $\begingroup$ If we can bound the finite sum $\sum_{m=1}^{n}\sin(m^k), \forall n\in \mathbb{N}$, then we can use Dirichlet test to prove the convergence. $\endgroup$ Jun 7, 2013 at 5:28

1 Answer 1


This is a replacement for my previous answer. The sum converges, and this fact needs even more math than I believed before.

Begin by using summation by parts. This gives $$\sum_{n=1}^N \left(\sum_{m=1}^N \sin(m^k) \right) \left( \frac{1}{n}-\frac{1}{n+1}\right) + \frac{1}{N+1} \left(\sum_{m=1}^N \sin(m^k) \right).$$ Write $S_n:= \left(\sum_{m=1}^n \sin(m^k) \right)$. So this is $$\sum_{n=1}^N S_n/(n(n+1)) + S_N/(N+1).$$ The second term goes to zero by Weyl's polynomial equidistribution theorem. So your question is equivalent to the question of whether $\sum s_n/(n(n+1))$ converges. We may as well clean this up a little: Since $|S_n| \leq n$, we know that $\sum S_n \left( 1/n(n+1) - 1/n^2 \right)$ converges. So the question is whether $$\sum \frac{S_n}{n^2}$$ converges.

I will show that $S_n$ is small enough that $\sum S_n/n^2$ converges absolutely.

The way I want to prove this is to use Weyl's inequality. Let $p_i/q_i$ be an infinite sequence of rational numbers such that $|1/(2 \pi) - p_i/q_i| < 1/q_i^2$. Such a sequence exists by a standard lemma. Weyl inequality gives that $$S_N = O\left(N^{1+\epsilon} (q_i^{-1} + N^{-1} + q_i N^{-k})^{1/2^{k-1}} \right)$$ for any $\epsilon>0$.

Thanks to George Lowther for pointing out the next step: According to Salikhov, for $q$ sufficiently large, we have $$|\pi - p/q| > 1/q^{7.60631+\epsilon}.$$ Since $x \mapsto 1/(2x)$ is Lipschitz near $\pi$, and since $p/q$ near $\pi$ implies that $p$ and $q$ are nearly proportional, we also have the lower bound $|1/(2 \pi) - p/q|> 1/q^{7.60631+\epsilon}$.

Let $p_i/q_i$ be the convergents of the continued fraction of $1/(2 \pi)$. By a standard result, $|1/(2 \pi) - p_i/q_i| \leq 1/(q_i q_{i+1})$. Thus, $q_{i+1} \leq q_i^{6.60631 + \epsilon}$ for $i$ sufficiently large. Thus, the intervals $[q_i, q_i^{7}]$ contain all sufficiently large integers.

For any large enough $N$, choose $q_i$ such that $N^{k-1} \in [q_i, q_i^7]$. Then Weyl's inequality gives the bound $$S_N = O \left( N^{1+\epsilon} \left(N^{-(k-1)/7} + N^{-1} + N^{-1} \right)^{1/2^{k-1}}\right)$$

So $$S_N = \begin{cases} O(N^{1-(k-1)/(7\cdot 2^{k-1}) + \epsilon}) &\mbox{ if } \ k\leq 7, \\ O(N^{1-1/(2^{k-1})+\epsilon}) &\mbox{ if } \ k\geq 8, \end{cases}$$ which is enough to make sure the sum converges. ${ }{}{}{}{}$

  • 2
    $\begingroup$ I hate to say this, but I am getting worried about whether I can push Weyl's inequality to do what I want. Weyl's inequality says, if $|\chi - p/q| < 1/q^2$, with $GCD(p,q)=1$, then $S_N = O(N^{1+\epsilon} (q^{-1} + N^{-1} + q N^{-d})^{1/2^d})$. What if the convergents to $\chi$ grow so fast that either $1/q$ or $q N^{-d}$ is as large as $1/ \log N$, for any choice of $q$? I was working from my notes from a course which claimed the $O(N^{1-1/2^d+\epsilon})$ bound, but I no longer follow the proof which was written there. So I have some concerns. $\endgroup$ Aug 17, 2010 at 15:38
  • $\begingroup$ David, your proof can be fixed up to work perfectly. Just needs one extra ingredient - pi has finite irrationality measure $\endgroup$ Aug 19, 2010 at 18:19
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    $\begingroup$ In fact, the mathworld website has a table of bounds for the irrationality measures of some constants, including pi. (mathworld.wolfram.com/IrrationalityMeasure.html) $\endgroup$ Aug 19, 2010 at 18:39
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    $\begingroup$ Perhaps I am just addled, but it seems to me that $\sum_{n=1}^N \left(\sum_{m=1}^N \sin(m^k) \right) \left( \frac{1}{n}-\frac{1}{n+1}\right) + \frac{1}{N+1} \left(\sum_{m=1}^N \sin(m^k) \right)=\left(1-\frac{1}{N+1} \right )S_N + \frac{1}{N+1}S_N = S_N$, not $\sum_{n=1}^N S_n/(n(n+1)) + S_N/(N+1)$ as claimed (I "fixed" $s_N$ to $S_N$). $\endgroup$
    – robjohn
    Aug 1, 2011 at 22:26
  • 1
    $\begingroup$ @robjohn The expression should be $ \sum_{n=1}^N\left(\sum_{m=1}^{\fbox{$n$}}\sin(m^k)\right)\cdots $, which gives the result David announced. $\endgroup$
    – jathd
    Jan 25, 2014 at 22:59

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