Let $K$ be a local field of characteristic $0$ with valuation $v$. I think 
$$\lim\_{\substack{q\in K\\q\to1}}\sum\_{n\ge0}\prod\_{j=1}^n\frac{q^j-1}{q-1}$$ converges to $\sum\_{n\ge0}n!\in K$ but I did not manage to prove it. Is my guess correct and if yes, can I have a hint or a proof of this fact?


Thanks in advance.



Convergence of a $p$-adic series
Is this question being asked just out of curiosity? As far as I know, the series $\sum\_{n \geq 0} n!$ is not important in $p$-adic analysis.


For $j \in \mathbf N$ and $q \not= 1$, let $(j)\_q = (q^j-1)/(q-1) = 1 + q + \cdots + q^{j-1}$. As $q \rightarrow 1$ we have $(j)\_q \rightarrow j$ so set $(j)\_1 = j$ for $j \in \mathbf N$. Set $(n)!\_q = (n)\_q(n-1)\_q\cdots (2)\_q(1)\_q$ for $n \in \mathbf Z^+$ and $(0)!\_q = 1$. Then $(n)!\_1 = n!$. You're looking at $\sum\_{n \geq 0} (n)!\_q$ as $q \rightarrow 1$ in $K$ and you want to show this series tends to $\sum\_{n \geq 0} n! = \sum\_{n \geq 0} (n)!\_1$ as $q$ tends to $1$, of course after we know that $\sum\_{n \geq 0} (n)!\_q$ converges in $K$ for $q$ close to $1$. It will turn out "close" can mean $|q-1| < 1$.


For $j \in \mathbf N$ and $|q-1| < 1$, show $|(j)\_q - j| \leq |q-1|$, so in the ring $\mathcal O\_K/(q-1)$, $(j)\_q \equiv j$ for all $j \in \mathbf N$ and thus $(n)!\_q \equiv n!$ for all $n \in \mathbf N$. Therefore 
$$
\left|\sum\_{n = 0}^N (n)!\_q - \sum\_{n = 0}^N n!\right| = \left|\sum\_{n = 0}^N \left((n)!\_q - n!\right)\right| \leq |q-1|.
$$ 
Since $K$ is a local field we know $n! \rightarrow 0$ in $K$ as $n \rightarrow \infty$. *If* we knew $(n)!\_q \rightarrow 0$ as $n \rightarrow \infty$ too then we could pass to the infinite series and say 
$$
\left|\sum\_{n \geq 0} (n)!\_q - \sum\_{n \geq 0} n!\right| = \left|\sum\_{n \geq 0} \left((n)!\_q - n!\right)\right| \leq |q-1|.
$$
Then let $q \rightarrow 1$ and you'd have what you were asking about. Does $(n)!\_q \rightarrow 0$ in $K$ as $n \rightarrow \infty$?


The field $K$ contains some $\mathbf Q\_p$. For that prime $p$, if $|q-1| < (1/p)^{1/(p-1)}$ and $q \not= 1$ then check (using $p$-adic exponential and logarithm, or other methods) that $|(j)\_q| = |j|$ for all $j \in \mathbf N$; it's obvious at $q=1$. Then $|(n)\_q!| = |n!|$ for all $n \in \mathbf N$, so $(n)!\_q \rightarrow 0$ as $n \rightarrow \infty$ when $|q-1| < (1/p)^{1/(p-1)}$, which is good enough since such $q$ are a neighborhood of $1$ and you're interested in a limit as $q \rightarrow 1$. 


But actually we have $(n)!\_q \rightarrow 0$ as $n \rightarrow \infty$ under the weaker condition $|q-1| < 1$. For such $q$ we can't use the $p$-adic exponential and logarithm and we won't have $|(j)\_q| = |j|$ all the time, but that's okay. For all $j$ we have $|(j)\_q| \leq 1$, so $|(n)!\_q| \leq |(p^r)\_q|$ where $p^r \leq n < p^{r+1}$. Since $n \rightarrow \infty$ is the same as $r \rightarrow \infty$, it suffices to show $(p^r)\_q \rightarrow 0$ in $K$ as $r \rightarrow \infty$ when $|q-1| < 1$. At $q = 1$ as have $(p^r)\_q = p^r$, which clearly tends to $0$ in $K$ as $r \rightarrow \infty$, so let $0 < |q-1| < 1$. Then our task is the same as showing $q^{p^r} \rightarrow 1$ in $K$ as $r \rightarrow \infty$, and that is a special case of the continuity of the function $q^x = \sum\_{k \geq 0} (q-1)^k\binom{x}{k}$ for $x \in \mathbf Z\_p$ (as $x \rightarrow 0$).


For $|q-1| < 1$ in $K$ we can think of the series $S(q) := \sum\_{n \geq 0} (n)!\_q$ as a function of $q$. The OP asked if $S(q)$ is continuous at $q=1$, and we saw it is. It's of course natural, having seen that, to ask if $S(q)$ is continuous in $q$ for all $q$ with $|q-1| < 1$. Let's compare $S(q\_1)$ to $S(q\_2)$ when $|q\_1 - 1| < 1$ and $|q\_2-1| < 1$. 


For $j \in \mathbf Z^+$, the difference $(j)\_{q\_1} - (j)\_{q\_2} = \sum\_{k=0}^{j-1} (q\_1^k - q\_2^k)$ is in $(q\_1-q\_2)\mathbf Z[q\_1,q\_2]$, so $|(j)\_{q\_1} - (j)\_{q\_2}| \leq |q\_1-q\_2|$. Therefore in the ring $\mathcal O\_K/(q\_1-q\_2)$ we have $(j)\_{q\_1} \equiv (j)\_{q\_2}$ for all $j \in \mathbf Z^+$ (at $j = 0$ too, but that's irrelevant here), so $(n)!\_{q\_1} \equiv (n)!\_{q\_2}$ for all $n \in \mathbf Z^+$ and also obviously at $n = 0$. Thus $|(n)!\_{q\_1} - (n)!\_{q\_2}| \leq |q\_1-q\_2|$ for all $n \in \mathbf N$, so 
$$
|S(q\_1) - S(q\_2)| = \left|\sum\_{n \geq 0} (n)!\_{q\_1} - \sum\_{n \geq 0} (n)!\_{q\_2}\right| \leq |q\_1-q\_2|, 
$$
which shows $S(q)$ is uniformly continuous in $q$ for $|q-1| < 1$ in $K$.


