# Convergence of a $p$-adic series

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?

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 $$K$$.

• While maybe not of the most importance, Ram does have a paper concerning this series and others like it. @KConrad likely knew of that already and maybe it doesn't go terribly far, but it at least is still actively cited. – lemiller Jun 3 '19 at 21:29
• @lemiller your link is paywalled (or at least requires some permission to access). It would be good to have a link available to all. Did you mean the article mast.queensu.ca/~murty/padic.pdf? – KConrad Jun 5 '19 at 0:19
• Ah yes, I did, thanks! The link I have was to the mathscinet review, showing the continued citations. I should have clarified. – lemiller Jun 5 '19 at 14:39