# Inequality with e, factorial, exponential function...

• Dec 9th 2012, 01:11 AM
wilhelm
Inequality with e, factorial, exponential function...
Hi!

Could you help me prove this inequality?
$\displaystyle e^{n-1} \cdot n! < n^{n+1}$

When I try induction , I get:
$\displaystyle e^n \cdot (n+1)! = e^{n-1} \cdot n! \cdot e \cdot (n+1)<$

and this is when I have no idea how to get $\displaystyle (n+1)^{n+2}$. Could I ask for a small hint? It cannot be that hard to prove :)
Or maybe you have another, more constructive way to prove this inequality?
• Dec 9th 2012, 02:29 AM
topsquark
Re: Inequality with e, factorial, exponential function...
Quote:

Originally Posted by wilhelm
Hi!

Could you help me prove this inequality?
$\displaystyle e^{n-1} \cdot n! < n^{n+1}$

When I try induction , I get:
$\displaystyle e^n \cdot (n+1)! = e^{n-1} \cdot n! \cdot e \cdot (n+1)<$

and this is when I have no idea how to get $(n+1)^{n+2}$. Could I ask for a small hint? It cannot be that hard to prove :)
Or maybe you have another, more constructive way to prove this inequality?

-Dan
• Dec 9th 2012, 02:38 AM
MarkFL
Re: Inequality with e, factorial, exponential function...
Dan, I have found the inequality is correct for $\displaystyle 1<n\in\mathbb{N}$ and provided a proof by induction on another site.

For example, for $\displaystyle n=2$, we have:

$\displaystyle 2e<8$

and for $\displaystyle n=3$ we have:

$\displaystyle 6e^2<3^4$

-Mark
• Dec 9th 2012, 03:21 AM
topsquark
Re: Inequality with e, factorial, exponential function...
Quote:

Originally Posted by MarkFL2
Dan, I have found the inequality is correct for $\displaystyle 1<n\in\mathbb{N}$ and provided a proof by induction on another site.

For example, for $\displaystyle n=2$, we have:

$\displaystyle 2e<8$

and for $\displaystyle n=3$ we have:

$\displaystyle 6e^2<3^4$

-Mark

I worked out a few on my calculator for convenience and used "e" instead of Exp. (Doh)

Thanks for the catch.

-Dan