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?

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 [latex](n+1)^{n+2}[/latex]. 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?

Made an oopsie! (Headbang)

-Dan

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

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