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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?

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    $\begingroup$ "$\lim_{n \to \infty} \frac{2^n}{n!}=0$" This is only true in real numbers. In $2$-adic numbers, this series doesn't converge (which is something you need to prove) $\endgroup$ – Wojowu Feb 28 '16 at 18:54
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    $\begingroup$ Do you know how to calculate $v(n!)$? $\endgroup$ – Bruno Joyal Feb 28 '16 at 19:03
  • $\begingroup$ @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. $\endgroup$ – S Blank Feb 28 '16 at 19:09
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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Feb 28 '16 at 22:49
  • $\begingroup$ 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 $\Bbb Q_2$. $\endgroup$ – Lubin Mar 1 '16 at 5:13
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$\nu_{2}(\frac{2^{n}}{n!})$ 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 $4$.

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  • $\begingroup$ Is there a direct way of showing $V_p(\frac{2^n}{n!})$ is 1 when n is a power of 2? $\endgroup$ – S Blank Mar 1 '16 at 21:03
  • $\begingroup$ To figure out the V_2 function, you subtract the terminal 0's in n! from those in 2^n, both in base 2. Other comments have given the formula for the terminal 0's in n!. For 2^n you have just n terminal 0's. Put in n = 2^m for any whole number m and find the difference. You'll be amazed, I was. $\endgroup$ – Oscar Lanzi Mar 1 '16 at 21:35
  • $\begingroup$ I used the geometric series when $n=2^k$ and it all worked out fine! Thanks. $\endgroup$ – S Blank Mar 1 '16 at 21:43

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