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

Is there a way to assess the convergence of the following series?

From numerical estimations it seems to be convergent but I don’t know how to prove it.

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 $$2πe2\pi e$$ is a rational number with a prime numerator. We first prove the following claims:

Lemma 1. If $$pp$$ is an odd prime number and $$S⊂ZS\subset \mathbb Z$$ so that
$$∑s∈Se2πis/p∈R,\sum_{s\in S}e^{2\pi i s/p}\in\mathbb R,$$
then $$∑s∈Ss≡0modp\sum_{s\in S}s\equiv 0\bmod p$$.

Proof. Let $$ζ=e2πi/p\zeta=e^{2\pi i/p}$$. We have
$$∑s∈Sζs=∑s∈Sζ−s,\sum_{s\in S}\zeta^s=\sum_{s\in S}\zeta^{-s},$$
since the sum is its own conjugate. As a result, since the minimal polynomial of $$ζ\zeta$$ is $$ζp−1ζ−1\frac{\zeta^p-1}{\zeta-1}$$, we see
$$xp−1x−1|∑s∈S(xp+s−xp−s),\frac{x^p-1}{x-1}\bigg|\sum_{s\in S}\left(x^{p+s}-x^{p-s}\right),$$
where we have placed each element of $$ss$$ in $$[0,p)[0,p)$$. The polynomial on the left is coprime with $$x−1x-1$$ and the polynomial on the right has it as a factor, so
$$xp−1x−1|∑s∈S(xp+s−1+⋯+xp−s).\frac{x^p-1}{x-1}\bigg|\sum_{s\in S}\left(x^{p+s-1}+\cdots+x^{p-s}\right).$$
Now, the quotient of these two polynomials must be an integer polynomial, so in particular the value of the left-side polynomial at $$11$$ must divide the value of the right-side polynomial at $$11$$. This gives $$p|∑s∈S2s,p|\sum_{s\in S}2s,$$ finishing the proof.

Define
$$an=n∑k=0n!k!.a_n=\sum_{k=0}^n \frac{n!}{k!}.$$

Lemma 2. If $$pp$$ is a prime number,
$$p−1∑n=0an≡−1modp.\sum_{n=0}^{p-1}a_n\equiv -1\bmod p.$$
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*}\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 $$j=n-kj=n-k$$. The inside sum is a sum of a polynomial over all elements of $$\mathbb Z/p\mathbb Z\mathbb Z/p\mathbb Z$$, and as a result it is $$00$$ as long as the polynomial is of degree less than $$p-1p-1$$ and it is $$-1-1$$ for a monic polynomial of degree $$p-1p-1$$. Since the only term for which this polynomial is of degree $$p-1p-1$$ is $$j=p-1j=p-1$$, we get the result.

Now, let $$2\pi e = p/q2\pi e = p/q$$. Define $$\mathcal E(x)=e^{2\pi i x}\mathcal E(x)=e^{2\pi i x}$$ to map from $$\mathbb R/\mathbb Z\mathbb R/\mathbb Z$$, and note that $$\mathcal E(x+\epsilon)=\mathcal E(x)+O(\epsilon)\mathcal E(x+\epsilon)=\mathcal E(x)+O(\epsilon)$$. 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*}\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 $$\frac{qe(n+p)!}{p}\frac{qe(n+p)!}{p}$$ “modulo $$11$$.” 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*}\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*}\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 $$mm$$ is the remainder when $$nn$$ is divided by $$pp$$. The terms with $$j>mj>m$$ in this sum go to $$00$$, giving us
$$\sum_{j=0}^m \frac{m!}{(m-j)!}=a_m.\sum_{j=0}^m \frac{m!}{(m-j)!}=a_m.$$
Putting this together, we see that
$$\sin((n+p)!)=\operatorname{Im}\mathcal E\left(\frac{qa_{n\bmod p}}p\right)+O\left(\frac 1n\right).\sin((n+p)!)=\operatorname{Im}\mathcal E\left(\frac{qa_{n\bmod p}}p\right)+O\left(\frac 1n\right).$$
In particular, the convergence of our sum would imply, since the $$O(1/n)O(1/n)$$ terms give a convergent series when multiplied by $$O(1/n)O(1/n)$$, that
$$x_N=\operatorname{Im}\sum_{n=1}^N\frac 1n\mathcal E\left(\frac{qa_{n\bmod p}}p\right)x_N=\operatorname{Im}\sum_{n=1}^N\frac 1n\mathcal E\left(\frac{qa_{n\bmod p}}p\right)$$
should converge. In particular, $$\{x_{pN}\}\{x_{pN}\}$$ must converge, which implies that
$$\sum_{m=0}^{p-1}\mathcal E\left(\frac{qa_m}p\right)\sum_{m=0}^{p-1}\mathcal E\left(\frac{qa_m}p\right)$$
must be real (as otherwise the series diverges like the harmonic series). By Lemma 1, this implies that
$$\sum_{m=0}^{p-1}a_m=0\bmod p,\sum_{m=0}^{p-1}a_m=0\bmod p,$$