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 $\beta$.

  • $\begingroup$ What does "finite irrationality measure" mean? $\endgroup$
    – Igor Rivin
    Dec 8, 2013 at 15:37
  • $\begingroup$ Irrationality measure: en.wikipedia.org/wiki/… $\endgroup$ Dec 8, 2013 at 15:38
  • $\begingroup$ Ah, that explains it, thnx. $\endgroup$
    – Igor Rivin
    Dec 8, 2013 at 15:59
  • 1
    $\begingroup$ $\sin x<x\iff\sin(\sin x)<\sin x$. (You can verify this geometrically, by noticing that the arc x is larger than the chord, which itself is larger than the sine, as the former is the hypotenuse, and the latter is the leg). Then use the fact that $\displaystyle\sum_{n=1}^\infty\frac{\sin n}n$ converges to $\dfrac{\pi-1}2$, which you can prove by using either Euler's formula, or the alternating series test. $\endgroup$
    – Lucian
    Dec 25, 2013 at 2:22
  • 2
    $\begingroup$ Achtung, @Lucian: $\sin(\sin x)<\sin(x)$ only if $\sin x> 0$, i.e. only if $\left\lfloor\frac{x}{\pi}\right\rfloor$ is even! $\endgroup$ Dec 28, 2013 at 13:10

3 Answers 3


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.

  • 1
    $\begingroup$ You have written "This gives: $$\left|\sum_{n=1}^{+\infty}\frac{\sin(\sin n)}{n}\right|\leq K\sum_{k=1}^{+\infty}\frac{k}{k^3},$$ for istance." - How did you get that? Also, note that the ratio$$\frac{\sigma_{k+1}(x)}{\sigma_k(x)}$$ can be negative. $\endgroup$ Dec 28, 2013 at 22:39
  • $\begingroup$ Simply write $\sin(\sin n)$ as a Fourier series and use the fact that $\sum_{n=1}^{+\infty}\frac{\sin(kn)}{n}=\frac{\pi-k}{2}=O(k)$. Moreover, $\frac{\sigma_{k+1}(x)}{\sigma_{k}(x)}$ cannot be negative on $[0,\pi/2]$ since $\frac{\sin x}{x}$ is non-negative on $[0,\pi/2]$. $\endgroup$ Dec 28, 2013 at 22:56
  • $\begingroup$ @Jack This is really cool; it took me a moment, though, to flesh out the core assertion you make at the beginning of your answer, namely, $\sum_{n\geq1}f(n)/n$ converges whenever $f$ is an odd periodic function in $C^4(\mathbb R)$. Correct me if I'm wrong, but it seems like you're suggesting we write $f(x) = \sum_{k\geq1}c_k\sin{(kx)}$, where $c_k=o(k^{-3})$, and then interchange the order of summation: $\sum_{n\geq 1} f(n)/n = \sum_{k\geq1}c_k\sum_{n\geq1}{\sin{(kn)}\over n}$. The change is justified because $\sum_{n\geq1}{\sin{(nx)}\over n}$ converges uniformly. Is this what you had in mind? $\endgroup$ Dec 29, 2013 at 1:04
  • $\begingroup$ @Nick Strehlke: exactly. The next step is to prove that: $$\mathcal{S}_k=\sum_{n=1}^{+\infty}\frac{\sigma_k(n)}{n}=-\frac{1}{2}+O\left(k^{-1/2}\log k\right)$$ by using the inequality $\sigma_{k}(x)\leq\min\left(\sin(x),\sqrt{\frac{3}{k}}\right)$ that is also very nice. $\endgroup$ Dec 29, 2013 at 2:31
  • 1
    $\begingroup$ @Jack Actually, I just realized that $\sum_{n\geq1}{\sin{(nx)}\over n}$ doesn't converge uniformly (it converges to a discontinuous function, namely, $-\operatorname{arg}(1-e^{ix})$). Nor does $\sum_{n\geq1}{\sin{(nk)}\over n}$ converge absolutely when $k$ is an integer. So how do we justify the change in order of summation that leads to $\sum_{n\geq 1}f(n)/n = \sum_{k\geq 1}c_k\sum_{n\geq1} {\sin{(kn)}\over n}$? $\endgroup$ Dec 29, 2013 at 18:40

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).


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.

  • 1
    $\begingroup$ Actually, it converges for every $k$, due to the fact that the measure of irrationality of $\frac{1}{2\pi}$ is finite. See mathoverflow.net/questions/150863/…. $\endgroup$ Dec 8, 2013 at 18:05
  • $\begingroup$ I must be dense -- I don't see how the answer to this question follows from the one you are pointing to. $\endgroup$
    – Igor Rivin
    Dec 8, 2013 at 19:53
  • $\begingroup$ It follows since $f(x)=\sigma_k(x)$ is $C^\infty$, $2\pi$-periodic and of zero average, and $1/2\pi$ has finite irrationality measure. Therefore, $s_n=\sigma_k(1)+\sigma_k(2)+\cdots+\sigma_k(n)$ is a bounded sequence, for every $k$, due to mathoverflow.net/questions/150863/…, and by Abel summation method, the above series converges. $\endgroup$ Dec 10, 2013 at 22:24
  • $\begingroup$ I think you misunderstand the question. $k$ is fixed, while in the Bruijn's book $k\to\infty$. $\endgroup$
    – vesszabo
    Dec 27, 2013 at 13:36
  • $\begingroup$ @vesszabo no, I did not misunderstand. $\endgroup$
    – Igor Rivin
    Dec 27, 2013 at 14:24

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.