# Convergence of $\sum_{n=1}^{\infty} \frac{\sin(n!)}{n}$

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.

• For 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". Apr 28, 2018 at 18:39
• @GabrielRomon Yes It is. en.m.wikipedia.org/wiki/Carleson%27s_theorem Apr 29, 2018 at 22:23
• I'm thinking use multiple angle formula for the sine and maybe Dirichlet test.
– MKu
Dec 3, 2018 at 8:32
• Can 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? Jun 3, 2019 at 19:37
• @SpencerKraisler You can do that, but your "upper bound" is not an absolutelly convergent series, so that would serve no purpose Jul 10, 2019 at 15:38

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 $$\frac{\zeta^p-1}{\zeta-1}$$, we see $$\frac{x^p-1}{x-1}\bigg|\sum_{s\in S}\left(x^{p+s}-x^{p-s}\right),$$ where we have placed each element of $$s$$ in $$[0,p)$$. The polynomial on the left is coprime with $$x-1$$ and the polynomial on the right has it as a factor, so $$\frac{x^p-1}{x-1}\bigg|\sum_{s\in S}\left(x^{p+s-1}+\cdots+x^{p-s}\right).$$ Now, the quotient of these two polynomials must be an integer polynomial, so in particular the value of the left-side polynomial at $$1$$ must divide the value of the right-side 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, $$\sum_{n=0}^{p-1}a_n\equiv -1\bmod p.$$ Proof. \begin{align*} \sum_{n=0}^{p-1}a_n &=\sum_{0\leq k\leq n\leq p-1}\frac{n!}{k!}\\ &=\sum_{0\leq n-k\leq n\leq p-1}(n-k)!\binom n{n-k}\\ &=\sum_{j=0}^{p-1}\sum_{n=j}^{p-1}n(n-1)\cdots(n-j+1)\\ &\equiv \sum_{j=0}^{p-1}\sum_{n=0}^{p-1}n(n-1)\cdots(n-j+1)\pmod p, \end{align*} where we have set $$j=n-k$$. 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 $$p-1$$ and it is $$-1$$ for a monic polynomial of degree $$p-1$$. Since the only term for which this polynomial is of degree $$p-1$$ is $$j=p-1$$, 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}^{p-1}\frac{(n+p)!}{(n+p-j)!} &=\sum_{j=0}^{p-1}(n+p)(n+p-1)\cdots(n+p-j+1)\\ &\equiv \sum_{j=0}^{p-1}m(m-1)\cdots (m-j+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 $$\sum_{j=0}^m \frac{m!}{(m-j)!}=a_m.$$ 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 $$\sum_{m=0}^{p-1}\mathcal E\left(\frac{qa_m}p\right)$$ must be real (as otherwise the series diverges like the harmonic series). By Lemma 1, this implies that $$\sum_{m=0}^{p-1}a_m=0\bmod p,$$ which contradicts Lemma 2.