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
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    $\begingroup$ 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 $\endgroup$
    – Conrad
    Commented Jul 28, 2021 at 14:47
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    $\begingroup$ Above of course $x_n=5 \times 9\times .....(4n+1) \pi/2$ to get all those $\sin x/k=1$ for $k=1,5,9,.....$ $\endgroup$
    – Conrad
    Commented Jul 28, 2021 at 15:03
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    $\begingroup$ @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? $\endgroup$
    – fedja
    Commented Jul 28, 2021 at 19:58
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    $\begingroup$ @fedja neat approach and the result on the reciprocal of primes is not that hard imho so I think the solution works very nicely $\endgroup$
    – Conrad
    Commented Jul 28, 2021 at 21:14

2 Answers 2


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.

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


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