Convergence of $\sum \limits_{n=1}^{\infty}\sin(n^k)/n$

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?

Answer

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.
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Attribution
Source : Link , Question Author : curious , Answer Author : Sungjin Kim

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