How to $2^n/n! \leq 4/n$ by induction I'm trying to prove that $\frac{2^n}{n!} \leq \frac{4}{n}$ for all positive integers $n$.
It's obvious this holds for $n=1$.
My inductive hypothesis is that $\frac{2^n}{n!} \leq \frac{4}{n}$.
I want to show that $\frac{2^{n+1}}{(n+1)!} \leq \frac{4}{(n+1)}$.
I can see that $\frac{2^{n+1}}{(n+1)!} = \frac{2^n \cdot 2}{n! \cdot (n+1)} \leq 
\frac{4 \cdot 2}{n(n+1)}$, where the inductive hypothesis is applied in the last step.
All I need to show is that $\frac{4 \cdot 2}{n(n+1)} \leq \frac{4}{n+1}$ and I'm done. But I can't think of a good justification for this.
Any hints or suggestions?

$\frac{4\cdot 2}{n(n+1)} = \frac{2}{n}\frac{4}{n+1}\leq \frac{4}{n+1}$, since $\frac{2}{n}\leq 1$ for $n\geq 2$.
In the case $n=1$, $2/1 \leq 4/1$.
