# $\exp(2)$ does not converge $2$-adically.

We have $\exp(2)= \sum_{i=0}^n {\frac{2^n}{n!}}$.

I am trying to show that $\exp(2)$ does not converge $2$-adically.

i.e. I need to show $\nu_2 (\frac{2^n}{n!})$ does not tend to $\infty$ as $n\to \infty$, where $\nu_p (x) = \max$ { $a : p^a$divides $x$ }.

$\nu_p (x)$ is $\infty$ only when $x$ is $0$.

However, since $\lim_{n \to \infty} \frac{2^n}{n!}=0$, I'll have $\nu_2 (\frac{2^n}{n!})\to \infty$ and hence $\exp(2)$ does converge $2$-adically, which is the exact opposite of what I'm trying to prove.

What went wrong there?

• "$\lim_{n \to \infty} \frac{2^n}{n!}=0$" This is only true in real numbers. In $2$-adic numbers, this series doesn't converge (which is something you need to prove) – Wojowu Feb 28 '16 at 18:54
• Do you know how to calculate $v(n!)$? – Bruno Joyal Feb 28 '16 at 19:03
• @Bruno Joyal: I know that $V_p(n!) = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor$ and so on. – S Blank Feb 28 '16 at 19:09
• See Wikipedia or locally for example this thread. The $p$-adic exponential series $\exp (x)$ converges only when $|x|_p<p^{-1/(p-1)}$. This follows from the formula for $\nu_p(n!)$ that you know. For large $n$ the r.h.s. is approximately $n/(p-1)$, so $x^n$ must be divisible by a power of $p$ that grows faster than $n/(p-1)$. Therefore $x$ needs to be divisible by a higher power than $p^{1/(p-1)}$. Those fractions come into play in extensions of the $p$-adic field. – Jyrki Lahtonen Feb 28 '16 at 22:49
• If $\exp(2)$ converged in $\Bbb Q_2$, then there would be an element $\alpha\in\Bbb Q_2$ such that $\log\alpha=2$. There are such things in the algebraic closure of $\Bbb Q_2$, in fact infinitely many of them, but none is in $\Bbb Q_2$. – Lubin Mar 1 '16 at 5:13

$\nu_{2}(\frac{2^{n}}{n!})$ does not go to infinity as n goes to infinity. The function is unbounded, true, but it is not monotonically increasing with $n$. It drops when $n$ hits a power of $2$ and gives you a lot of factors of $2$ in the factorial. In fact $\nu_{2}(\frac{2^{n}}{n!})$ drops all the way down to $1$ when $n$ is a power of $2$.
The exponential function converges in $2$-adics when the argument is a multiple of $4$.
• Is there a direct way of showing $V_p(\frac{2^n}{n!})$ is 1 when n is a power of 2? – S Blank Mar 1 '16 at 21:03
• I used the geometric series when $n=2^k$ and it all worked out fine! Thanks. – S Blank Mar 1 '16 at 21:43