# $(\frac{n}{e})^{n} < n! < (\frac{n}{e} + n\varepsilon)^{n}$ doesn't comply with the limit definition?

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?

• 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. – Calvin Godfrey Jan 10 at 20:39
• @CalvinGodfrey, they just mean $e$. – Joe Jan 10 at 20:40
• @Joe In that case $\exp(1)$ would be more accurate. – cansomeonehelpmeout Jan 10 at 20:41
• @cansomeonehelpmeout, yes that would have been more appropriate. I'm just explaining what OP meant. I absolutely do not condone their notation. – Joe Jan 10 at 20:43
• @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 – dpd Jan 10 at 21:02

## 1 Answer

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 $$\epsilon$$.