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

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

Inspiration : I lately found out that $S_1$ converges. I assume $S_2$ merges by the indispensable examination. Was the inquiry recognized as a whole?

2019-05-09 11:18:05
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Answers: 1

This is a substitute for my previous solution. The amount merges, and also this reality requires a lot more mathematics than I thought in the past.

Begin by utilizing summation by components. This offers $$\sum_{n=1}^N \left(\sum_{m=1}^N \sin(m^k) \right) \left( \frac{1}{n}-\frac{1}{n+1}\right) + \frac{1}{N+1} \left(\sum_{m=1}^N \sin(m^k) \right).$$ 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 2nd term mosts likely to absolutely no by Weyl's polynomial equidistribution theorem. So your inquiry amounts the inquiry of whether $\sum s_n/(n(n+1))$ merges. We might too tidy this up a little : Since $|S_n| \leq n$ , we understand that $\sum S_n \left( 1/n(n+1) - 1/n^2 \right)$ merges. So the inquiry is whether $$\sum \frac{S_n}{n^2}$$ merges.

I will certainly show that $S_n$ is tiny sufficient that $\sum S_n/n^2$ merges definitely.

The means I intend to confirm this is to make use of Weyl's inequality. Allow $p_i/q_i$ be a boundless series of sensible numbers such that $|1/(2 \pi) - p_i/q_i| < 1/q_i^2$ . Such a series exists by a typical lemma. Weyl inequality considers that $$S_N = O\left(N^{1+\epsilon} (q_i^{-1} + N^{-1} + q_i N^{-k})^{1/2^{k-1}} \right)$$ for any kind of $\epsilon>0$ .

Many Thanks to George Lowther for mentioning the next action : According to Salikhov, for $q$ completely huge, we have $$|\pi - p/q| > 1/q^{7.60631+\epsilon}.$$ Since $x \mapsto 1/(2x)$ is Lipschitz near $\pi$ , and also given that $p/q$ near $\pi$ indicates that $p$ and also $q$ are virtually symmetrical, we additionally have actually the lower bound $|1/(2 \pi) - p/q|> 1/q^{7.60631+\epsilon}$ .

Allow $p_i/q_i$ be the convergents of the proceeded portion of $1/(2 \pi)$ . By a typical outcome, $|1/(2 \pi) - p_i/q_i| \leq 1/(q_i q_{i+1})$ . Hence, $q_{i+1} \leq q_i^{6.60631 + \epsilon}$ for $i$ completely huge. Hence, the periods $[q_i, q_i^{7}]$ have all completely huge integers.

For any kind of huge adequate $N$ , pick $q_i$ such that $N^{k-1} \in [q_i, q_i^7]$ . After that Weyl's inequality offers 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 suffices to see to it the amount merges. ${ }{}{}{}{}$

2019-05-10 05:44:22

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