Convergence of series in p-adic norm

Question

I want to know the convergence of the series

$\sum_{n\geq 0} {{p^{n+1}}\choose{p^n}}$

My idea is to show the convergence of the partial sum $|s_n-s_{n-1}|_p$ but I am stucked in expressing the terms properly.

Is there any hint to this question?

Answer 1

Let me tell you a cute identity for binomial coefficients: $$ \binom{a}{b} = \frac{a}{b}\binom{a-1}{b-1}. $$ Therefore $$ \binom{p^{n+1}}{p^n} = p\binom{p^{n+1}-1}{p^n-1}. $$

Now let me tell you a cute congruence for binomial coefficients modulo $p$, going back to Kummer: $$ \binom{a_0 + a_1p + \cdots + a_rp^r}{b_0 + b_1p + \cdots + b_rp^r} \equiv \binom{a_0}{b_0}\binom{a_1}{b_1}\cdots \binom{a_r}{b_r} \bmod p $$ where $a_j$ and $b_j$ are base $p$ digits. The base $p$ digits of $p^{n+1}-1$ and $p^n-1$ are all $p-1$, so $\binom{p^{n+1}-1}{p^n-1} \bmod p$ can be computed as follows: $$ \binom{(p-1) + \cdots + (p-1)p^{n-1} +(p-1)p^n}{(p-1) + \cdots + (p-1)p^{n-1} + (0)p^n} \equiv \binom{p-1}{p-1} \cdots \binom{p-1}{p-1}\binom{p-1}{0} \bmod p, $$ so $\binom{p^{n+1}-1}{p^n-1} \equiv 1 \bmod p$.

Since $\binom{p^{n+1}-1}{p^n-1}$ is not divisible by $p$, ${\rm ord}_p\binom{p^{n+1}}{p^n} = {\rm ord}_p(p\binom{p^{n+1}-1}{p^n-1}) = 1$. Therefore every term in the series you wrote about has the same $p$-adic absolute value, so the series does not converge (terms do not tend to $0$).

Answer 2

The sum converges in the p-adics if and only if we have $\lim_{n \to \infty} \binom{p^{n+1}}{p^n} = 0$ like you say. Using Legendre's formula we can get an exact answer for what the terms of the series tend to,

$$v_p \left( \binom{p^{n+1}}{p^n}\right) = v_p\left(\frac{p^{n+1}!}{p^n! (p^{n+1}-p^n)!}\right) = v_p (p^{n+1}!) - v_p(p^n!) - v_p((p^{n+1}-p^n)!)$$

$$=\frac{p^{n+1} - s_p(p^{n+1})}{p-1} - \frac{p^n-s_p(p^n)}{p-1} - \frac{p^{n+1}-p^n-s_p(p^{n+1}-p^n)}{p-1}$$

$$=\frac{- s_p(p^{n+1}) + s_p(p^n) + s_p(p^{n+1}-p^n)}{p-1}$$

Now since a power of $p$ is simply a single $1$ followed by $0$s the first two terms are easy to evaluate, and since $p^{n+1}-p^n = (p-1)p^n$ this is already written in base $p$, since it is the digit $p-1$ followed by $n$ following $0$s.

$$=\frac{- 1 + 1 + p-1 }{p-1} = 1$$

Because the valuation is the constant $1$, it must be that,

$$\left|\binom{p^{n+1}}{p^n}\right|_p = p^{-1} \not \to 0$$

So it does not converge.

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There seemed to be some confusion about how it converges from reading the comments, hopefully I can give some analogies to series of real numbers for how this fails to converge to help clear that up.

Because your series is bounded the entire time, it resembles the way in which the real series $\sum_{n \ge 0} (-1)^n$ fails to converge, because it is always bounded with its partial sums perpetually "meandering around".

Keep in mind we still have sums that "diverge to infinity" here in the p-adics, in the sense that the partial sums get arbitrarily large. For instance the p-adic series $\sum_{n \ge 0} \frac{1}{p^n}$ has its terms $\left|\frac{1}{p^n}\right|_p = p^n \to \infty$ which is how the real series $\sum_{n \ge 0} n$ also fails to converge.