Is the series $\sum_{i=k}^\infty\frac{\sin\left(\frac{x}{i}\right)}{i}$ bounded?

Comparison with $$\dfrac{1}{i^2}$$ shows that the series $$\sum_{i=k}^\infty\frac{\sin\left(x/i\right)}{i}$$ is convergent. However, the function of $$x$$ thus obtained appears to be bounded, with the bound approaching zero in $$k$$. I have no idea how to prove this, every method I know gives me no better a bound than the obvious $$\dfrac{\pi^2x}{6}$$. My suspicion is that prior to reaching $$\left|\dfrac{x}{i}\right|<1$$, the values of $$\dfrac{x}{i}$$ modulo $$2\pi$$ have an asymptotic distribution which helps similar to other identities with alternating signs bounded by $$\dfrac{1}{i}$$.

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– Pedro
Commented Jul 16, 2021 at 8:09
• by splitting at $x$ one can easily get a $C \log x$ bound but I suspect the function is unbounded with at least some $x_n \to \infty$ for which $|f(x_n)| >> \log \log x_n$ since for something like $x_n=5 \times 9\times .....(4n+1)$ the series has a large initial part until $\log x_n$ which I think can be inferior bounded by $C \log \log x_n$; from $x^{\epsilon}$ on one can easily bound the series, but between $\log x$ and $x^{\epsilon}$ the oscillations may or may not cancel out the initial segment; for numeric checks, I would try an $x_n$ as above at very large values if possible Commented Jul 28, 2021 at 14:47
• Above of course $x_n=5 \times 9\times .....(4n+1) \pi/2$ to get all those $\sin x/k=1$ for $k=1,5,9,.....$ Commented Jul 28, 2021 at 15:03
• @Conrad I tried to implement your strategy though I needed a slight modification of it, which forced me to use the Dirichlet theorem on primes in the arithmetic progressions. Do you see how one could possibly avoid it? Commented Jul 28, 2021 at 19:58
• @fedja neat approach and the result on the reciprocal of primes is not that hard imho so I think the solution works very nicely Commented Jul 28, 2021 at 21:14

It is unbounded though to prove it, I need the fact that the sum of reciprocals of the primes $$p=4k+1$$ is infinite, which seems like a bit too heavy tool for this problem.

The first observation is that for every $$M,q,\varphi$$, we have $$\left|\frac 1M\sum_{m=0}^{M-1}\sin(2\pi mq+\varphi) \right|\le\min(1,\tfrac1{M\|q\|})$$ where $$\|q\|$$ is the distance from $$q$$ to the nearest integer (just write $$\sin\theta=\Im e^{i\theta}$$ and sum the corresponding geometric progression).

The second observation is that for $$0\le x\le B$$, we have $$\left|\sum_{k\ge B}\frac{\sin(x/k)}{k}\right|\le\sum_{k\ge B}\frac B{k^2}\le 2\,.$$

Now enumerate the primes $$p=4k+1$$ as $$p_1,p_2,\dots$$ and put $$A=\prod_{j=1}^N p_j$$. Consider $$x_m=(4m+1)A\frac\pi 2$$, $$m=0,\dots,M-1$$.

Denote the sum of the series by $$S(x)$$. We want to look at the average $$\frac 1M\sum_{m=0}^{M-1} S(x_m)$$. First of all, we can truncate the series in the definition of $$S(x_m)$$ to the first $$B=2\pi AM$$ terms (the rest is uniformly bounded).

Now if $$k\not\mid A$$, then by our first observation the average of the corresponding term in the series is in absolute value at most $$\frac 1k\frac 1{M\|\frac Ak\|}$$. However, $$\|\frac Ak\|\ge \frac 1k$$ for $$k\le 2A$$ and equals $$\frac Ak$$ for $$k>2A$$. Thus, the sum of these averages is at most $$\sum_{k\le 2A}\frac 1M+\sum_{2A\le k\le B}\frac 1{AM}\le \frac{2A}M+\frac{2\pi AM}{AM}\le 2(1+\pi)\,,$$ provided that $$M>A$$, say.

Thus we are left with the terms for which $$k\mid A$$. However in this case $$A/k\equiv 1\mod 4$$, so $$\sin(x_m/k)=1$$ for all $$m$$ and all these terms push in the same direction (positive), which makes the corresponding sum at least $$\sum_{k\mid A}\frac 1k\ge\sum_{j=1}^N \frac 1{p_j}\,,$$ which can be made arbitrarily large.

• Miraculous answer. Commented Jul 28, 2021 at 22:58
• Very nice! ${}{}$ Commented Jul 29, 2021 at 3:18
• I don't think $\sum_{p \equiv 1 \text{ mod 4}} 1/p = +\infty$ is that heavy. See writeup below. Commented Aug 4, 2021 at 21:19
• @mathworker21 Yep, this particular case allows a rather short proof, I agree :-) Thanks for spelling out the details. Commented Aug 4, 2021 at 22:07

Here is a writeup of Fedja's solution with proofs of lemmas and required machinery included.

Lemma $$1$$: $$\sum_{p \equiv 1 \text{ mod 4}} 1/p = +\infty$$.

Proof: For $$\chi$$ the dirichlet character mod $$4$$ ($$\chi(n) = 0$$ if $$2 \mid n$$ and $$\chi(n) = (-1)^{(n-1)/2}$$ otherwise), we have $$\log(\sum_{n=1}^\infty \frac{\chi(n)}{n^s}) = \log(\prod_p \frac{1}{1-\chi(p)p^{-s}}) = \sum_p \sum_{m \ge 1} \frac{\chi(p)^m}{p^{ms}} = \sum_p \frac{\chi(p)}{p^s}+O(1)$$ for real $$s$$ near $$1$$. Since $$\sum_p 1/p = +\infty$$, if $$\sum_{p \equiv 1 \text{ mod 4}} 1/p < +\infty$$, then the RHS approaches $$-\infty$$ as $$s \downarrow 1$$ while the LHS approaches $$\log(1-1/3+1/5-1/7+\dots)$$, a contradiction since $$1-1/3,1/5-1/7,\dots > 0$$. $$\square$$

Lemma $$2$$: For any $$\theta,\phi \in \mathbb{R}$$ and any $$M \ge 1$$, we have $$\sum_{m=0}^{M-1} \sin(2\pi m \theta+\phi) \le 1/\|\theta\|_{\mathbb{R}/\mathbb{Z}}$$.

Proof: We wish to prove $$Im[e^{i\phi}\sum_{m=0}^{M-1} e^{2\pi i m \theta}] \le 1/\|\theta\|_{\mathbb{R}/\mathbb{Z}}$$, so it suffices to show $$|\sum_{m=0}^{M-1} e^{2\pi i m\theta}| \le 1/\|\theta\|_{\mathbb{R}/\mathbb{Z}}$$. The magnitude of the geometric sum is $$|\frac{e^{2\pi i M\theta}-1}{e^{2\pi i \theta}-1}| \le \frac{2}{|e^{2\pi i \theta}-1|}$$. It suffices to restrict attention to $$\theta \in [-1/2,1/2]$$, and here $$|e^{2\pi i \theta}-1| \ge 2|\theta|$$ finishes the job. $$\square$$

Now onto the problem. Let $$f(x) = \sum_{k=1}^\infty \sin(x/k)/k$$.

Claim: There are arbitrarily large $$x$$ with $$f(x) = \Omega(\log\log\log x)$$.

Proof: Take $$N \ge 1$$. Let $$A = \prod_{j=1}^N p_j$$, where $$p_1,p_2,\dots$$ are the primes congruent to $$1$$ mod $$4$$. Let $$M \ge 1$$ be a parameter (just take $$M=A$$) and $$B = \lceil 2\pi M A\rceil$$. Let $$x_m = 2\pi m A+A\frac{\pi}{2}$$ for $$0 \le m \le M-1$$. We show that $$\frac{1}{M}\sum_{m=0}^{M-1} f(x_m) = \Omega(\log\log N)$$, which will finish the proof by pigeonhole and that each $$x_m \le 10A^2 \le \exp(30N)$$. We start with $$\left|\frac{1}{M}\sum_{m=0}^{M-1} \sum_{k > B} \frac{\sin(x_m/k)}{k}\right| \le \frac{1}{M}\sum_{m=0}^{M-1} x_m\frac{2}{B} \le 4.$$ Then, using Lemma $$2$$, we do $$\left|\frac{1}{M}\sum_{m=0}^{M-1} \sum_{\substack{k \le B \\ k \not \mid A}} \frac{\sin(x_m/k)}{k}\right| = \left|\sum_{\substack{k \le B \\ k \not \mid A}} \frac{1}{k}\frac{1}{M}\sum_{m=0}^{M-1} \sin(2\pi\frac{A}{k}m+\frac{A\pi}{2k})\right| \le \sum_{\substack{k \le B \\ k \not \mid A}} \frac{1}{k}\frac{1}{M}\frac{1}{\|A/k\|_{\mathbb{R}/\mathbb{Z}}},$$ which we can trivially bound by $$\sum_{k \le 2A} \frac{1}{k}\frac{1}{M}\frac{1}{1/k} + \sum_{2A < k \le B} \frac{1}{k}\frac{1}{M}\frac{1}{A/k} = \frac{2A}{M}+\frac{B}{AM} \le 10.$$ Therefore, using that $$A/k \equiv 1 \text{ mod 4}$$ if $$k \mid A$$, we see $$\frac{1}{M}\sum_{m=0}^{M-1} f(x_m) \ge \frac{1}{M}\sum_{m=0}^{M-1} \sum_{k \mid A} \frac{\sin(x_m/k)}{k} - 14 = \sum_{k \mid A} \frac{1}{k}-14 = \Omega(\log\log N),$$ where we used a quantitative form of Lemma $$1$$ (which actually also follows from the given proof). $$\square$$

Remaining Questions:

Claim: For any large $$x > 0$$, we have $$f(x) = O(\log x)$$.

Proof: $$|\sum_{k \le x} \sin(x/k)/k| \le \sum_{k \le x} 1/k = O(\log x)$$ and $$|\sum_{k > x} \sin(x/k)/k| \le x\sum_{k > x} 1/k^2 \le 2$$. $$\square$$

Question $$1$$: What's the best lower bound achieved for arbitrarily large $$x$$? Do we have $$f(x) = \Omega(\log \log x)$$ for arbitrarily large $$x$$? How about $$\Omega(\log x)$$?

I find the following question interesting.

Question $$2$$: Defining $$g(R) := \sup_{x > 0} \sum_{k=1}^R \sin(x/k)$$ what is $$\limsup_{R \to \infty} g(R)/R$$? It's at least $$\approx .578188$$ (the maximum of $$\sin(x)-x\text{Ci}(x)$$).

Finally, Fedja's proof does not resolve the following question.

Question $$3$$: Is $$\sum_{k=1}^\infty \frac{\sin(x/k)}{k\log k}$$ bounded as a function of $$x$$?