How to show $\epsilon > \frac{3^n -1}{n! + 124} > \frac{1}{n!}$

Question

I'm having some problems with proving infimum of some set. I got stuck where I need to show the following:

$$ \forall_{\epsilon > 0}\ \exists_{n \in \mathbb{N}}\ \epsilon > \frac{3^n -1}{n! + 124} $$

So obviously $\epsilon > \frac{3^n -1}{n! + 124} > \frac{1}{n!} $ (I guess for $n \geq 5 $)

But how to move it from this point?

EDIT:

So following your hints:

$n! + 124 = 1\cdot 2\cdot 3 \cdot 4 \cdots n + 124 > 1\cdot 2\cdot 3\cdot 4\cdot 4\cdots 4= 6\cdot 4^{n-3}$

$\epsilon > \frac{3^n}{4^{n-3}} > \frac{3^n -1}{6 \cdot 4^{n-3}} > \frac{3^n -1}{n! + 124}$

So I have $\frac{\epsilon}{64} > (\frac{3}{4})^n$

So $ \log_{}{\frac{\epsilon}{64} > n \cdot \log{}{\frac{3}{4}}}$

So $n > \log_{\frac{3}{4}}{\frac{\epsilon}{64}}$

So there exists $ n = floor(\log_{\frac{3}{4}}{\frac{\epsilon}{64}}) + 1$

Is this correct?

Answer 1

The important point of confusion and that has to be corrected is that you don't want to show that $$ \varepsilon > \frac{3^n-1}{n!+124}>\frac{1}{n!}, $$ since that would not show your statement. It goes in the wrong direction in some sense.

Rather, you should try to find some convincing $f(n)$ such that $$ \varepsilon > f(n) > \frac{3^n-1}{n!+124} $$ for large $n$, and such that $f(n)$ is decreasing to zero as $n\to+\infty$.

I think @5xum gives you a nice way to do this with $$ f(n)=\frac{3^n-1}{6\cdot 4^{n-3}}. $$

Answer 2

Hint:

For $n > 3$, you can use the fact that $$n! + 124 = 1\cdot 2\cdot 3 \cdot 4\cdots n + 124 > 1\cdot 2\cdot 3\cdot 4\cdot 4\cdots 4> 6\cdot 4^{n-3} $$

Also, use the fact that if $a>b>0$, then $0<\frac 1a < \frac1b$