Is the P-adic limit $\lim_{n \to \infty} p^{p^n}$ 0 or 1?

Question

I was playing around making up some limits of things I thought would be interesting to check out and came to something I thought was quite simple on the surface but found it wasn't very consistent with other things in my mind.

$$\lim_{n \to \infty} p^{p^n} = 0$$

To me this seemed very obviously 0, since more and more times you multiply p by itself you really get smaller and smaller with respect to the p-adic metric. But something about the concept of exponents and "counting" just doesn't seem to make sense in the context of p-adic numbers since there is no ordering.

In talking about a similar limit for the p-adic logarithm I've heard it mentioned that the exponent would be getting smaller and smaller, so perhaps it's just as fine to interpret this as,

$$\lim_{n \to \infty} p^{p^n} = p^0 = 1$$

Which is perhaps a bit dubious at first, but if I were to expand this as the Mahler series,

$$p^x = \sum_{n=0}^\infty (p-1)^n \binom{x}{n} = 1 + (p-1)x + (p-1)^2\frac{x(x-1)}{2} + \cdots$$

Then I run into this issue of taking the limit from this definition a bit more seriously,

$$\lim_{n \to \infty} 1 + (p-1)p^{n} + (p-1)^2\frac{p^{n}(p^{n}-1)}{2} + \cdots$$

$$=1$$

However maybe I'm doing some sort of circular reasoning or even running into simply some kind of choice of convention here, I'm not quite so sure.

Answer 1

Your first thought was exactly correct: the limit is $0$. You are getting confused here about the domain of exponentiation: $p^n$ is only defined when $n$ is an integer, not when $n$ is a $p$-adic number. So you shouldn't be surprised if this function ends up being ill-behaved when you think of $n$ as a $p$-adic number instead. In particular, the function $n\mapsto p^n$ is not continuous as a function $\mathbb{Z}\to\mathbb{Q}_p$ when you give $\mathbb{Z}$ the $p$-adic topology, so you cannot compute the limit of $p^{p^n}$ by taking the limit of the exponent as you have done.

(You might try defining $p^x$ for $p$-adic $x$ using the $p$-adic exponential and logarithm by $p^x=\exp(x\log p)$. However, this does not work since $p$ is not in the domain of the $p$-adic logarithm.)

Answer 2

This seems to be a classic case of unjustified interchange of summation and limits. In truth, the Mahler series has only finitely many (non-zero) terms for any given $n$, in fact when $x$ is a power of $p$ I think it only has two: the $n=0$ term and the $n=x$ term. But it is not necessarily the case that the limit of the sums is equal to the sum of the limits. Your series is highly analogous to the following sequence of infinite sums:

$$ 1 + -1 + 0 + 0 + 0 +\cdots, \\ 1 + 0 + -1 + 0 + 0 + \cdots, \\ 1 + 0 + 0 + -1 + 0 + \cdots, \\ 1 + 0 + 0 + 0 + -1 + \cdots, \\ \vdots. $$

It is obvious that in each row, the sum is $0$, so the sequence is identically $0$ (and thus clearly converges to $0$). However, if you instead take the limit of each term, the first column converges to $1$ and all other columns converge to $0$, so the sum of the limits is $1$.

Answer 3

I hope I’m not adding to your confusion by saying this:

In your expression $p^{\text{something}}$, the base, $p$, is a $p$-adic integer, but the exponent “something” is just an ordinary counting number, a genuine integer (though it can be negative as well). It is definitely not a $p$-adic number. So you have to think of what you’ve written as “$p^{\text{huge}}$”, thus an expression with limit zero.

EDIT:What I originally wrote below was careless and wrong, since I interchanged $t$ and $z$ in the formula.
And now I’m really going to undo everything and confuse you totally. For $p$-adic numbers of form $u=1+pa$, where $a$ can be any $p$-adic number, in other words, $\vert u-1\vert_p<1$, you may raise to an exponent that is not an ordinary integer, but in fact a $p$-adic integer in $\Bbb Z_p$. That is, you can define $(1+pa)^z$ for $a,z\in\Bbb Z_p$. Just use the Binomial Theorem that you learned in Calculus, $$ (1+t)^z=1+zt+\frac{z(z-1)}{2!}t^2+\frac{z(z-1)(z-2)}{3!}t^3+\cdots\, $$ and verify that with $t=pa$ and $z$ as I specified, the whole mess is $p$-adically convergent. This takes some work, but the expression is continuous in both $a$ and $z$, just as you had been hoping in the other, worse, situation.