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?

  • 1
    $\begingroup$ "$\lim_{n \to \infty} \frac{2^n}{n!}=0$" 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

$\nu_{2}(\frac{2^{n}}{n!})$ 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 $4$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.