# Stuck with showing for $x \in \mathbb{Z}_p$, $\sum_{n=1}^{\infty} \frac{x(x-1) \ldots (x-n+1)}{n!}$ converges in $\mathbb{Z}_p$.

I want to show that for arbitrary $x \in \mathbb{Z}_p$, $\sum_{n=1}^{\infty} \frac{x(x-1) \ldots (x-n+1)}{n!}p^{n}$ converges to an element in $\mathbb{Z}_p$. So, I wanted to use $\sum_{n\geq 0 } a_n$ converges iff $(a_n)_n$ converges to $0$. For this, I use, if $n = \sum_{i=0}^r a_i p^i$ is the $p$-adic extension of $n$ and $s_p(n) = \sum_{i=0}^r a_i$, then $ord_p(n!) = \frac{n - s_p(n)}{p-1}$. Therefore, I get:

$| \frac{x(x-1) \ldots (x-n+1)}{n!}p^{n} | = |\frac{x!}{n! (x-n)!}|p^{-n} = p^{-\frac{x - s_p(x)}{p-1} + \frac{n - s_p(n)}{p-1} + \frac{x-n - s_p(x-n)}{p-1} - n} = p^{\frac{s_p(x) - s_p(n) - s_p (x-n) }{p-1} - n}$ Now I am stuck with showing that this goes to $0$. How do you show this?

## 1 Answer

For $p>2$, your formula for $ord_p(n)$ shows that it is at most $\frac n{p-1}$, and the difference $n-\frac n{p-1}$ goes to infinity. We can ignore the rest of the numerator.

For $p=2$, the factors $n!$ and $p^n$ give $ord_2$ of at least $0$. Now as $n\to\infty$ the factor $x(x-1)\cdots$ contains more and more factors $2$; at least $n/2-1$.