Is there a way to assess the convergence of the following series? $$\sum_{n=1}^{\infty} \frac{\sin(n!)}{n}$$ From numerical estimations it seems to be convergent but I don't know how to prove it.

Are there infinitely many summands that are $\gt\frac{1}{2n}$? I don't think it converge..Does this help?https://math.stackexchange.com/questions/2227574/convergenceofsumn1inftyfracsinnn?rq=1 – Tony Ma Apr 27 '18 at 12:40

1If there is some $\epsilon>0$ such that $\sum_{n=1}^N \sin(n!)=O(N^{1\epsilon})$, then the series can be shown to converge by partial summation. I expect that $\sin(n!)$ is positive about as often as it is negative, and that $\sum_{n=1}^N\sin(n!)$ grows slowly, but I cannot see how to prove the necessary bound. – Julian Rosen Apr 27 '18 at 14:32

1Also, Wolfram shows that the partial sums will all be greater than 1 eventually, so the series must diverge. Also, doesnt $\frac{1}{n} \le \frac{\sin(n!)}{n} \le \frac{1}{n}$ get you any further? – Ramanujan Apr 28 '18 at 09:01

30For almost all $\theta\in [0,2\pi)$, the series $$\sum_{n=1}^{\infty} \frac{\sin(n! \theta)}n $$ converges. However, it is a completely different story if $\theta=1$ belongs to the "almost all". – Sungjin Kim Apr 28 '18 at 18:39

@i707107 is that a result from the theory of Fourier series ? – Gabriel Romon Apr 29 '18 at 07:26

2@GabrielRomon Yes It is. https://en.m.wikipedia.org/wiki/Carleson%27s_theorem – Sungjin Kim Apr 29 '18 at 22:23

3I'm thinking use multiple angle formula for the sine and maybe Dirichlet test. – MKu Dec 03 '18 at 08:32

4Can someone explain why we can't show that $\left  \frac{\sin(n!)}{n} \right  \leq \left  \frac{(1)^n}{n} \right$ for all $n \in \mathbb{N}$, implying the series is "bounded" by the alternating harmonic series? – Spencer Kraisler Jun 03 '19 at 19:37

@SpencerKraisler the RHS is equal to $\frac{1}{n}$. Without the absolute values, you actually have "$>$" for $n$ odd and $<$ for $n$ even – Jakob W. Jun 08 '19 at 21:47

12@SpencerKraisler You can do that, but your "upper bound" is not an absolutelly convergent series, so that would serve no purpose – David Jul 10 '19 at 15:38

1It may help to know $$ \sum_{n=1}^{\infty} \frac{\sin{n}}{n} = \frac{\pi  1}{2} $$ @JulienRosen I expect you are correct about the sign of the $\sin$ term. It is most likely true in the above equality as well. – Jeremy Sep 25 '19 at 18:56

1It must converge because sin(n˚) is 0 if n is a multiple of 360 and all factorials are from 6!. – Dec 07 '19 at 14:45

1A similar question has an answer here: https://math.stackexchange.com/questions/677719/thelimitofsinn The only difficulty is that neither 0*infinity nor o*infinity are defined. – Dec 07 '19 at 14:52

2Related: See https://mathoverflow.net/questions/248199/conditionalconvergenceofsumngeq1fracsinpnn – Kevin.S Dec 08 '19 at 06:10

The fact that there are infinitely many negative and positive terms seems to be a way to go. We do know that if the terms are perfectly alternate like a $(1)^n$ it converges right? I think there is some sort of generalization of that – Rick Apr 24 '20 at 22:51

1@TalesRickPerche unfortunately we do not know that $n! \pmod{2 \pi}$ is equidistributed (or even dense) in [0, 2pi) so we cannot prove convergence. I think this is basically an open problem... – Thomas May 06 '20 at 09:26

1Has there ever been a question with so many deleted answers? – David P May 30 '20 at 19:11

2@Thomas Even equidistribution is not strong enough to prove convergence. I once asked a question on that and was given a counter example which divergences. See https://math.stackexchange.com/questions/3488494/doesequidistributionimplyconvergence – QC_QAOA Jun 20 '20 at 13:18

It may help that $x!\approx\sqrt{2\pi x}(\frac{x}{e})^x$. The ratio of the two approaches 1 as $x\rightarrow\infty$. (Stirling's formula) – Graviton Jul 21 '20 at 13:27

Perhaps Kolmogorov's ThreeSeries theorem or Kolmogorov Maximal Inequality will provide some inspiration. Certainly Byron Schmuland "Random Harmonic Series" has thought long and hard: http://www.stat.ualberta.ca/people/schmu/preprints/rhs.pdf – ShpielMeister Jul 24 '20 at 18:30

@SpencerKraisler I think it's hard to say much about the sign of $sin(n!)$ in general if $n!$ is assumed to be an angle in radian measure. It's highly improbable that it alternates in sign. – richard gayle Jul 24 '20 at 19:23

I know we can find an irrational number x in $(0,\pi)$ so that $sin(n!x)$ has same tail behavior as $sin(n!)$. – Ken.Wong Jul 31 '20 at 20:01
1 Answers
Here is a proof that the answer is (almost certainly) not provable using current techniques. We will prove that the series in fact diverges if $2\pi e$ is a rational number with a prime numerator. We first prove the following claims:
Lemma 1. If $p$ is an odd prime number and $S\subset \mathbb Z$ so that $$\sum_{s\in S}e^{2\pi i s/p}\in\mathbb R,$$ then $\sum_{s\in S}s\equiv 0\bmod p$.
Proof. Let $\zeta=e^{2\pi i/p}$. We have $$\sum_{s\in S}\zeta^s=\sum_{s\in S}\zeta^{s},$$ since the sum is its own conjugate. As a result, since the minimal polynomial of $\zeta$ is $\frac{\zeta^p1}{\zeta1}$, we see $$\frac{x^p1}{x1}\bigg\sum_{s\in S}\left(x^{p+s}x^{ps}\right),$$ where we have placed each element of $s$ in $[0,p)$. The polynomial on the left is coprime with $x1$ and the polynomial on the right has it as a factor, so $$\frac{x^p1}{x1}\bigg\sum_{s\in S}\left(x^{p+s1}+\cdots+x^{ps}\right).$$ Now, the quotient of these two polynomials must be an integer polynomial, so in particular the value of the leftside polynomial at $1$ must divide the value of the rightside polynomial at $1$. This gives $p\sum_{s\in S}2s,$ finishing the proof.
Define $$a_n=\sum_{k=0}^n \frac{n!}{k!}.$$
Lemma 2. If $p$ is a prime number, $$\sum_{n=0}^{p1}a_n\equiv 1\bmod p.$$ Proof. \begin{align*} \sum_{n=0}^{p1}a_n &=\sum_{0\leq k\leq n\leq p1}\frac{n!}{k!}\\ &=\sum_{0\leq nk\leq n\leq p1}(nk)!\binom n{nk}\\ &=\sum_{j=0}^{p1}\sum_{n=j}^{p1}n(n1)\cdots(nj+1)\\ &\equiv \sum_{j=0}^{p1}\sum_{n=0}^{p1}n(n1)\cdots(nj+1)\pmod p, \end{align*} where we have set $j=nk$. The inside sum is a sum of a polynomial over all elements of $\mathbb Z/p\mathbb Z$, and as a result it is $0$ as long as the polynomial is of degree less than $p1$ and it is $1$ for a monic polynomial of degree $p1$. Since the only term for which this polynomial is of degree $p1$ is $j=p1$, we get the result.
Now, let $2\pi e = p/q$. Define $\mathcal E(x)=e^{2\pi i x}$ to map from $\mathbb R/\mathbb Z$, and note that $\mathcal E(x+\epsilon)=\mathcal E(x)+O(\epsilon)$. We have \begin{align*} \sin((n+p)!) &=\operatorname{Im}\mathcal E\left(\frac{(n+p)!}{2\pi}\right)\\ &=\operatorname{Im}\mathcal E\left(\frac{qe(n+p)!}{p}\right). \end{align*} We will investigate $\frac{qe(n+p)!}{p}$ "modulo $1$." We see that \begin{align*} \frac{qe(n+p)!}{p} &=q\sum_{k=0}^\infty \frac{(n+p)!}{pk!}\\ &\equiv q\sum_{k=n+1}^\infty \frac{(n+p)!}{pk!}\pmod 1\\ &=O(1/n)+q\sum_{k=n+1}^{n+p}\frac{(n+p)!}{pk!}\\ &=O(1/n)+\frac qp\left[\sum_{k=n+1}^{n+p}\frac{(n+p)!}{k!}\pmod p\right]. \end{align*} Now, \begin{align*} \sum_{k=n+1}^{n+p}\frac{(n+p)!}{k!}=\sum_{j=0}^{p1}\frac{(n+p)!}{(n+pj)!} &=\sum_{j=0}^{p1}(n+p)(n+p1)\cdots(n+pj+1)\\ &\equiv \sum_{j=0}^{p1}m(m1)\cdots (mj+1)\pmod p, \end{align*} where $m$ is the remainder when $n$ is divided by $p$. The terms with $j>m$ in this sum go to $0$, giving us $$\sum_{j=0}^m \frac{m!}{(mj)!}=a_m.$$ Putting this together, we see that $$\sin((n+p)!)=\operatorname{Im}\mathcal E\left(\frac{qa_{n\bmod p}}p\right)+O\left(\frac 1n\right).$$ In particular, the convergence of our sum would imply, since the $O(1/n)$ terms give a convergent series when multiplied by $O(1/n)$, that $$x_N=\operatorname{Im}\sum_{n=1}^N\frac 1n\mathcal E\left(\frac{qa_{n\bmod p}}p\right)$$ should converge. In particular, $\{x_{pN}\}$ must converge, which implies that $$\sum_{m=0}^{p1}\mathcal E\left(\frac{qa_m}p\right)$$ must be real (as otherwise the series diverges like the harmonic series). By Lemma 1, this implies that $$\sum_{m=0}^{p1}a_m=0\bmod p,$$ which contradicts Lemma 2.
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