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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 $\beta$.

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  • $\begingroup$ 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
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    $\begingroup$ $\sin x<x\iff\sin(\sin x)<\sin x$. (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
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    $\begingroup$ 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

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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.

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    $\begingroup$ 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
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    $\begingroup$ @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
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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).

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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.

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    $\begingroup$ 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

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