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 S⊂Z so that

∑s∈Se2πis/p∈R,

then ∑s∈Ss≡0modp.

*Proof.* Let ζ=e2πi/p. We have

∑s∈Sζs=∑s∈Sζ−s,

since the sum is its own conjugate. As a result, since the minimal polynomial of ζ is ζp−1ζ−1, we see

xp−1x−1|∑s∈S(xp+s−xp−s),

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

xp−1x−1|∑s∈S(xp+s−1+⋯+xp−s).

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|∑s∈S2s, finishing the proof.

Define

an=n∑k=0n!k!.

**Lemma 2.** If p is a prime number,

p−1∑n=0an≡−1modp.

*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*