In connection with the strange behavious of a certain sum in Mathematica (https://mathematica.stackexchange.com/q/210849/16361) I suspected a possible divergence but I could not prove of disprove it.

Here's the question: is the sum

$$s_1=\sum_{k=1}^\infty \frac{\sin\left(k(k-1)\right)}{k}$$

convergent or divergent?

Similarly with

$$s_2=\sum_{k=1}^\infty \frac{\sin(k^2)}{k}$$

Numerical evidence (partial sums) seem to indicate convergence.

EDIT 07.12.19

Actually, the story began one step earlier: I considered this unanswered question Convergence of $\sum_{n=1}^{\infty} \frac{\sin(n!)}{n}$

$$s_3 =\sum_{k=1}^\infty \frac{\sin(k!)}{k}$$

and wanted to simplify it replacing $k!$ with something simpler.

  • 1
    $\begingroup$ Convergence would probably follow from the assertion that the sequence $\{k^2/2\pi\}_{k=1}^\infty$ is uniformly distributed modulo $1$, as long as the assertion was a little bit quantitative. This assertion is probably true but I didn't immediately find a source. $\endgroup$ Dec 7, 2019 at 2:16
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    $\begingroup$ I think you can argue as follows. By Weyl differencing, the partial sums $\sum_{k=1}^N \sin k^2$ and $\sum_{k=1}^N \sin k(k-1)$ grow like $C\sqrt{N}$. Then you can use summation by parts in each sum to get a sum of terms bounded by $Ck^{-3/2}$ which should then converge. $\endgroup$
    – Zarrax
    Dec 7, 2019 at 2:58
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    $\begingroup$ @Zarrax You don't want to make an answer ? No need to write all the details, just indicate that $|\sum_{n\le N} \exp(in^2)|^2 = \sum\sum \exp(i (n+h)^2-in^2)$ $\endgroup$
    – reuns
    Dec 7, 2019 at 3:14
  • $\begingroup$ Idea similar to Greg Martin's: compare the sum with a random sum and use Kolmogoroff's theorem to prove convergence (en.wikipedia.org/wiki/Kolmogorov%27s_three-series_theorem) $\endgroup$ Dec 7, 2019 at 8:27
  • $\begingroup$ No, the random series can't be compared. For a random variable $r \sim N(0,1)$ we have $E[\frac {\sin(2 \pi r)}{k}] = 0$ and so the sum also vanishes. This means that the correlation between numerator and denominator is essential. $\endgroup$ Dec 7, 2019 at 9:12

3 Answers 3


I realized that I couldn't be sure my argument worked, so I'll just leave this here to see if any number theorists can finish it. Basically, the situation is as follows. Suppose that we can show that there is a constant $A > 0$ such that for any sufficiently large positive integer $N$ there are integers $a$ and $q$ with $(a,q) = 1$ and ${N \over A} < q < AN$ such that we have $$\bigg|{1 \over 2\pi} - {a \over q}\bigg| \leq {1 \over q^2} \tag{0}$$

In that case, by the Weyl inequality for quadratic polynomials (see http://citeseerx.ist.psu.edu/viewdoc/download?doi= for a proof), one has $$\bigg|\sum_{k=1}^N e^{i(k^2 - k)}\bigg| \leq CN^{1/2} \log N$$ $$\bigg|\sum_{k=1}^N e^{ik^2}\bigg| \leq CN^{1/2}\log N$$ Taking imaginary parts gives $$\bigg|\sum_{k=1}^N \sin(k^2 - k)\bigg| \leq CN^{1/2} \log N \tag{1}$$ $$\bigg|\sum_{k=1}^N \sin k^2\bigg| \leq CN^{1/2}\log N$$

Next, summation by parts gives $$\sum_{k=1}^N {\sin(k^2 - k) \over k} = {\sum_{k=1}^N \sin(k^2 - k) \over N} $$ $$+ \sum_{k=2}^{N+1}\bigg(\sum_{l=1}^k \sin(l(l-1))\bigg){1 \over k(k-1)} \tag{2}$$ Taking limits as $N$ goes to infinity, the term ${\sum_{k=1}^N \sin(k^2 - k) \over N}$ in $(2)$ goes to zero. On the other hand, if one inserts $(1)$ into $a_k = \bigg(\sum_{l=1}^k \sin(l(l-1))\bigg){1 \over k(k-1)}$, one sees that $|a_k| \leq C\ln|k| |k|^{-{3 \over 2}}$. Hence the sum on the right in $(2)$ is absolutely convergent. Thus as $N$ goes to infinity this sum converges as well. Hence the overall sum $\sum_{k=1}^{\infty} {\sin(k^2 - k) \over k}$ is convergent.

A similar argument would work for $\sum_{k=1}^{\infty} {\sin k^2 \over k}$.

So the issue becomes whether or not $(0)$ holds. Also note you can replace the condition ${N \over A} < q < AN$ by ${1 \over A}N^{\epsilon} < q < AN^{2-\epsilon}$ for any $\epsilon > 0$ and an analogous argument will work. Maybe someone here is familiar enough with rational approximations to irrational numbers to say one way or another.

  • $\begingroup$ "the term $\sum_{i=1}^N\sin(k^2-k)/N$ in (2) goes to zero". I thought the upper-bound in (1) would be $log N$. $\endgroup$
    – mikado
    Dec 7, 2019 at 18:40
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    $\begingroup$ mikado I just typoed there, it should be $N^{1 \over 2}$ and not $N$ in $(1)$ so that term is bounded by $C{\log N \over N^{1/2}}$. $\endgroup$
    – Zarrax
    Dec 7, 2019 at 18:51
  • $\begingroup$ @ Zarrax Possibly related math.stackexchange.com/questions/3185881/… $\endgroup$ Dec 8, 2019 at 18:44
  • $\begingroup$ @Dr.WolfgangHintze The convergence of the sum and its value is strongly connected with the irrationality measure of $\pi$. This is discussed here. There is this many-times edited claim on arxiv, which I have some doubts of. $\endgroup$
    – yarchik
    Dec 8, 2019 at 19:01
  • $\begingroup$ Every algebraic number has irrationality measure 2 according to the Roth theorem. $\endgroup$
    – yarchik
    Dec 8, 2019 at 19:06

Instead of proving convergence, I'll try calculating the limit.

For some large positive integer $n$ we have $$ s_1 = \sum_{k=1}^{\infty} \frac{\sin(k(k-1))}{k} = \sum_{k=1}^n \frac{\sin(k(k-1))}{k} + \sum_{k=n+1}^{\infty} \frac{\sin(k(k-1))}{k} $$ The first term we can calculate exactly. For the second term we assume that $k(k-1)\mod2\pi$ is distributed uniformly in the interval $[0,2\pi)$ and so the sine functions can be replaced by independent random variables with zero mean and a variance of $\frac12$. We find the mean value $$ \langle s_1\rangle = \sum_{k=1}^n \frac{\sin(k(k-1))}{k} + \sum_{k=n+1}^{\infty} \frac{\langle\sin(k(k-1))\rangle}{k}= \sum_{k=1}^n \frac{\sin(k(k-1))}{k} $$ and the total variance $$ \text{Var}(s_1) = \sum_{k=n+1}^{\infty} \frac{\frac12}{k^2} = \frac12\psi'(n+1)=\frac{1}{2n}+O(n^{-2}). $$ We thus find the physics-style estimate $$ s_1 = \left[\sum_{k=1}^n \frac{\sin(k(k-1))}{k}\right] \pm \frac{1}{\sqrt{2n}} $$ for $n\to\infty$.

Using $n=5\times10^7$ terms in the first sum, I find $s_1=0.3128±0.0001$.

In the same way, $s_2=0.1667±0.0001$.

More generally, we can try to plot the function $$ f(z) = \sum_{k=1}^{\infty} \frac{\sin(k(k+z))}{k} $$ where $s_1=f(-1)$ and $s_2=f(0)$: this function seems to have a lot of structure,

enter image description here

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    $\begingroup$ @ Roman +1 That's really elegant. In my comment using the random procedure I missed your splitting of of the sum into low and high end. The physics-style argument is convincing for me (as a physicist) although strictly speaking Weyl does not guarantee randomness for integer coefficients of the polynom P(n) (see my comment to Greg Martin). $\endgroup$ Dec 8, 2019 at 8:44
  • $\begingroup$ @Dr.WolfgangHintze yes, strictly speaking the assumption of randomness is just an assumption, and my results may be quite wrong. I suspect they are alright though, and would be interested in seeing an example of a deviation. I'm a physicist too, and think of irrational numbers in the way used here, for practical purposes. $\endgroup$
    – Roman
    Dec 14, 2019 at 19:49
  • $\begingroup$ @ Roman The more I think about your approach the more I like it. You might perhaps also be interested in my attempt to attack the sum analytically math.stackexchange.com/a/3466679/198592 $\endgroup$ Dec 16, 2019 at 8:22
  • $\begingroup$ @Dr.WolfgangHintze your Fourier formula looks interesting but I haven't been able to evaluate it into a simple result. Maybe it just transfers the difficulty to a different spot: integrating this formula over intervals of length 2pi and then summing these results (to infinity). But a good idea for sure. $\endgroup$
    – Roman
    Dec 17, 2019 at 19:48

Here is my first attempt to calculate a sum related to $s_2$

$$s_{2a} = \sum_{k=1}^\infty \frac{e{^{i n^2}}}{n}\tag{1}$$

by tranforming it into an integral. I don't know if we gain much from this.

Writing the numerator as an explicit Fourier transform

$$e{^{i n^2}} = \frac{1}{\sqrt{2 \pi }}\int_{-\infty }^{\infty } \left(\frac{1}{2}+\frac{i}{2}\right) e^{-i n z-\frac{i z^2}{4}} \, dz\tag{2}$$

we can do the $n$-sum of the integrand leading to

$$g(z)=-\frac{1}{\sqrt{2 \pi }}\left(\frac{1}{2}+\frac{i}{2}\right) e^{-\frac{1}{4} \left(i z^2\right)} \log \left(1-e^{-i z}\right)\tag{3}$$

and the sum becomes

$$s_{2a} = \int_{-\infty }^{\infty } g(z) \, dz\tag{4}$$

EDIT 16.12.19

We can simplify the integral splitting the integration interval it into two parts from $-\infty$ to $0$ and from $0$ to $\infty$, collecting it in an integral from $0$ to $\infty$ and then substituting $z\to 2 \sqrt{t}$ to get

$$s_{2a}=-\frac{\left(1+i\right)}{2\sqrt{2 \pi }}\int_0^\infty \frac{ e^{-i t} \log \left(4 \sin ^2\left(\sqrt{t}\right)\right)} {\sqrt{t}}\,dt\tag{5}$$

The singularitity of the integrand at $t\to0$ is of the type $\frac{\log(t)}{\sqrt{t}}$ which is integrable, and the others at $t=k^2 \pi^2$ are of the type $\log|t-t_0|$ and hence are integrable a fortiori.


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