Proof by induction - inequality and factorials

[eiokjhgf eiokjhgf]

Hello,

I'm trying to prove (3^n)*(n!) < (2*n)! for any integer n >= 2.

So, the first part is easy:
(3^2)*(2!) < (2*2)!
9*2 < 4!
18 < 24

Now we suppose it's true for any (3^k)*(k!) < (2*k)! where k >= 2 and
have to prove it for k+1 so I was trying the following:

(3^k)*(k!) < (2*k)! /multiply everything with (k+1)
(3^k)*(k+1)(k!) < (k+1)*(2*k)!
(3^k)*(k+1)! < (k+1)(2*k)!

And now I wouldn't know how to continue and make it develop it for k+1

Any help is very appreciated,

Thank you so much for your time

[Paul Sperry]

In article
<aec62ced-9008-4ab2...@t13g2000yqn.googlegroups.com>,
eiokjhgf eiokjhgf <eiok...@gmail.com> wrote:

> Hello,
>
> I'm trying to prove (3^n)*(n!) < (2*n)! for any integer n >= 2.
>
> So, the first part is easy:
> (3^2)*(2!) < (2*2)!
> 9*2 < 4!
> 18 < 24
>
> Now we suppose it's true for any (3^k)*(k!) < (2*k)! where k >= 2 and
> have to prove it for k+1 so I was trying the following:

You want to assume (3^k)*(k!) < (2*k)! for some k >= 2 and show that
(3^(k + 1))*((k + 1)!) < (2*(k + 1))!.

Notice that 3^(k + 1) = 3*(3^k), (k + 1)! = (k + 1)*(k!) and
(2*(k + 1))! = (2k + 2)*(2k + 1)*((2k)!)

[...]

--
Paul Sperry
Columbia, SC (USA)

[eiokjhgf eiokjhgf]

Thank you very much :)