Convergence of a series with repeated sines

Question

Show that the series $$ \sum_{n=1}^\infty \frac{\sin\big(\sin(n)\big)}{n}, $$ converges.

More generally, show that for every $k\in\mathbb N$ the series $$ \sum_{n=1}^\infty \frac{\sigma_k(n)}{n}, $$ converges, where $\sigma_1(x)=\sin(x)$ and $\sigma_{k+1}(x)=\sin\big(\sigma_k(x)\big)$.

Note. I am looking for an elementary proof, if such is available. I do know that such elementary proof exists, for $k=2$, since that case was qualifying exam (University of Adelaide) a few years ago. On the other hand, I know of a non elementary proof (for $k$ general), using the nontrivial fact that $\dfrac{1}{2\pi}$ has finite irrationality measure. I imagine that any proof would have as a first step establishing the fact that the sequence $$ \sum_{j=1}^n \sigma_k(j), \quad n\in\mathbb N, $$ is bounded.

Update. Actually, using advanced tools (i.e., irrationality measure), it turns out that even $$ \sum_{n=1}^\infty \frac{\sigma_k(n)}{n^a}, $$ converges, for every $a>0$, and $k\in\mathbb N$, while the same tools can not determine whether $$ \sum_{n=1}^\infty \frac{\sin\big(\sin(\beta n)\big)}{n^a}, $$ converges, for every $\beta$.

Answer 1

To prove the convergence it is sufficient to show that $$\left|\sum_{n=1}^{+\infty}\frac{\sin(kn)}{n}\right|= \left|\frac{\pi-k}{2}+\pi \left\lfloor\frac{k}{2\pi}\right\rfloor\right| < \frac{\pi}{2} \tag{1}$$ by regarding the LHS as the imaginary part of a geometric series and by applying Abel's lemma.

Then we have that the Fourier coefficients of $\sin(\sin x)$ decay pretty fast since $\sin(\sin x)\in C^3([-\pi,\pi])$. If we are allowed to write $\sin(\sin x)$ as its Fourier series and to switch the sums (this is crucial) we have: $$\left|\sum_{n=1}^{+\infty}\frac{\sin(\sin n)}{n}\right|\leq K\sum_{k=1}^{+\infty}\frac{\pi}{k^2},$$ for istance. I am expecting that the same holds for $\sigma_k(x)$, since, in general, if $f(x)$ is an odd periodic function belonging to $C^{3}(\mathbb{R})$ and we are allowed to switch sums, $$\sum_{n=1}\frac{f(n)}{n}$$ converges because the $k$-th Fourier coefficient of $f(x)$ is $o(k^{-2})$ and $(1)$ holds.

By using Bessel functions we can write: $$\sin(\sin n)=\sum_{k=0}^{+\infty}2\cdot J_{2k+1}(1)\sin((2k+1) n),\tag{2}$$ where the convergence is uniform, but now we have to justify the sum-switch procedure. This is more-or-less the same as proving that the partial sums of $\sin(\sin n)$ are bounded. Exploiting $(2)$ we get: $$\sum_{n=0}^{N}\sin(\sin n)\leq\sum_{k=0}^{+\infty}\frac{2\cdot J_{2k+1}(1)}{|\sin(k+1/2)|},$$ and we have to prove that the RHS is bounded. If we define $\|x\|$ as the distance between $x$ and the closest integer, we have: $$\sum_{k=0}^{K}\frac{2\cdot J_{2k+1}(1)}{|\sin(k+1/2)|}\leq \sum_{k=0}^{K}\frac{1}{4^k(2k+1)!\cdot|\sin(k+1/2)|}\leq \sum_{k=0}^{K}\frac{1}{2^{2k+1}(2k+1)!\cdot\left|\frac{2k+1}{2\pi}\right|},$$ so, in order to prove that the partial sums of $\sin(\sin n)$ are bounded, it is sufficient to prove that there exists a positive number $C$ such that $$ \left|\frac{2k+1}{2\pi}\right|\geq\frac{C}{(2k+1)!},\tag{3} $$ holds for every $k$. This is much weaker than requiring that $2\pi$ has a finite irrationality measure: if the terms of the continued fraction of $2\pi$ do not grow too fast, since $$\frac{1}{(a_n+1)q_n^2}<\frac{1}{q_{n}(q_{n+1}+q_{n})}<\left|2\pi-\frac{p_n}{q_n}\right|<\frac{1}{q_n q_{n+1}}<\frac{1}{q_n^2}, $$ holds for every convergent $\frac{p_n}{q_n}$ of $2\pi=[a_0;a_1,a_2,\ldots]$, $(3)$ follows from the Legendre theorem.

Answer 2

If $f(x)$ is an odd continuous function with range $[-a,a]$, and $f(x) \leq |x|$, then $f^k(x)$ has the same property for all $k \geq 1$.

If a sequence $a_n$ has a symmetric (even) frequency distribution measure in the interval $[-a,a]$, then $f^k(a_n)$ has the same property.

Based on this, and taking $a_n$ to be the value of $(n \mod 2\pi)$ in $(-\pi,\pi)$, by Abel summation the $\sum \frac{f^k(a_n)}{n}$ converges if $s_n = f^k(a_1) + \dots + f^k(a_n)$ is $O(n^c)$ for $c < 1$. Uniform distribution modulo $1$ gives only $s_n = o(n)$ which is an epsilon less than what we need.

The difference $d_n = |\frac{s_n}{n} - \int_{-\pi}^{\pi} f^k(x) dx|$ converges to $0$ by uniform distribution (applied to $f^k \circ g$ for $g$ that re-coordinatizates the interval to make the distribution uniform). For $f^k$ of bounded variation, we are asking a special case of the discrepancy problem for sequence $a_n$ : is there a bound $d_n = O(n^{-u})$ for $u > 0$?

This power-of-$n$ improvement in the convergence is true for any sequence with a positive irrationality measure. Maybe a sledgehammer but it shows the problem does not need any special property of $f(x)=\sin x$ except that $f^k$ have bounded variation (which holds if $f'$ exists and is continuous).

Answer 3

I could be totally wrong, but it seems to me that the analysis in de Bruijn's book, p. 157(see, esp, a couple of pages down, like p. 159, where the $x$ dependence is discussed) would seem to indicate that the series would diverge for large $k,$ at least.