# mathematical induction again

• Jul 30th 2007, 10:15 AM
tottenc
mathematical induction again
I need to use mathematical induction to prove $\displaystyle 2^n<n!$ for all n>=4. The base case is easy, and in the inductive step I get to $\displaystyle 2(2^n)<(n+1)n!$ But am I going in the wrong direction, or am I just missing something?
• Jul 30th 2007, 10:41 AM
tukeywilliams
Assume $\displaystyle 2^k < k!$ for $\displaystyle k \geq 4$. Then $\displaystyle 2^{k+1} < 2k! < (k+1)! = k!(k+1)$. $\displaystyle k+1 > 2$ for $\displaystyle k >1$. Hence $\displaystyle 2^{k+1} < (k+1)!$ for $\displaystyle k \geq 4$ and we have proved the inductive step.
• Jul 30th 2007, 10:49 AM
tottenc
Sorry, stupid typo
$\displaystyle 2^n<n!$ for all
$\displaystyle n>=4$
• Jul 30th 2007, 11:16 AM
Plato
Assume that $\displaystyle N \ge 4\quad \& \quad 2^N < N!$ for the inductive step.

Then $\displaystyle 2^{N + 1} = 2\left( {2^N } \right) < \left( {N + 1} \right)\left( {N!} \right) = \left( {N + 1} \right)!$
• Jul 30th 2007, 05:14 PM
tottenc
The answer to these things is always staring me right in the face, and I just can't see it until someone points it out. Thanks again
• Jul 31st 2007, 04:34 AM
topsquark
Quote:

Originally Posted by tottenc
The answer to these things is always staring me right in the face, and I just can't see it until someone points it out. Thanks again

It just takes some practice. ;)

-Dan