I try to understand what I've overlooked, when I came up with this inequality:

First, we have this limit: $$\lim\limits_{n \to \infty} \sqrt[n]{\frac{n!}{n^n}} = \frac{1}{e}$$ Which gives, by the definition of limit and some simple transformations:

$\frac{1}{e} - \varepsilon < \sqrt[n]{\frac{n!}{n^n}} < \frac{1}{e} + \varepsilon$

$(\frac{n}{e} - n\varepsilon)^{n} < n! < (\frac{n}{e} + n\varepsilon)^{n}\quad\forall \varepsilon > 0$

Then, we have this well-known inequality (multiple proofs can be found on math.stackexchange): $$(\frac{n}{e})^{n} < n!$$ So we have: $$(\frac{n}{e})^{n} < n! < (\frac{n}{e} + n\varepsilon)^{n}$$ According to this inequality, we cannot make $\varepsilon$ arbitrary small, which contradicts the definition of limit. What am I missing here?

  • 1
    $\begingroup$ What do you mean by $\exp$? As far as I know, it's meant to be a function, but you're not giving it an input. $\endgroup$ – Calvin Godfrey Jan 10 at 20:39
  • $\begingroup$ @CalvinGodfrey, they just mean $e$. $\endgroup$ – Joe Jan 10 at 20:40
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    $\begingroup$ @Joe In that case $\exp(1)$ would be more accurate. $\endgroup$ – cansomeonehelpmeout Jan 10 at 20:41
  • $\begingroup$ @cansomeonehelpmeout, yes that would have been more appropriate. I'm just explaining what OP meant. I absolutely do not condone their notation. $\endgroup$ – Joe Jan 10 at 20:43
  • $\begingroup$ @mfl, thanks, I was stuck because I didn't consider that all 3 of expressions in this inequality are sequences and thought in terms of fixed n $\endgroup$ – dpd Jan 10 at 21:02

Actually yes, we can make $\epsilon$ arbitrarily small. Note that your argument is a limit argument, meaning it doesn't hold for every $n$. It only holds for all $n \geq N$, where $N$ depends on $\epsilon$.


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