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?
n=1sin(n!)n
From numerical estimations it seems to be convergent but I don’t know how to prove it.

Answer

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 2πe is a rational number with a prime numerator. We first prove the following claims:

Lemma 1. If p is an odd prime number and SZ so that
sSe2πis/pR,
then sSs0modp.

Proof. Let ζ=e2πi/p. We have
sSζs=sSζs,
since the sum is its own conjugate. As a result, since the minimal polynomial of ζ is ζp1ζ1, we see
xp1x1|sS(xp+sxps),
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
xp1x1|sS(xp+s1++xps).
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|sS2s, finishing the proof.


Define
an=nk=0n!k!.

Lemma 2. If p is a prime number,
p1n=0an1modp.
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.

Attribution
Source : Link , Question Author : Leonardo Massai , Answer Author : Carl Schildkraut

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