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本帖最后由 hbghlyj 于 2023-4-15 00:15 编辑 math.stackexchange.com/questions/2756038/convergence-of-sum-n-1-infty-frac-sinnn?noredirect=1
$$a_n=\sum_{k=0}^n \frac{n!}{k!}.$$
我们将证明$(a_n)$发散,如果$2πe$是一个有质数分子的有理数.但是以目前的技术不足以证明$2πe$为无理数.
引理 1. $p$为奇素数, $S\subset \mathbb Z$,使得$\displaystyle\sum_{s\in S}e^{2\pi i s/p}\in\mathbb R,$则$\displaystyle\sum_{s\in S}s\equiv 0\bmod p$.
证明.
Lemma 2. $p$是素数,则
$$\sum_{n=0}^{p-1}a_n\equiv -1\bmod p.$$
证明.
\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}\\
\xlongequal{j=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*}
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.
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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. |
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