Convergence of $\sum_{k=1}^\infty \frac{\sin(k(k-1))}{k}$

Question

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.

Answer 1

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=10.1.1.190.3347&rep=rep1&type=pdf 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.

Answer 2

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,

Answer 3

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.