# prove 3^n < n! for all n>=7

• Jun 18th 2010, 09:47 PM
atSydneyUni
prove 3^n < n! for all n>=7
Hi,

I'm familiar with induction with equal signs. We did it in high school. However I can't seem to understand out to prove by induction using the logical operator '<' (less than).

And i'm stuck on this question. All help is greatly appreciated.
• Jun 18th 2010, 09:59 PM
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Quote:

Originally Posted by atSydneyUni
Hi,

I'm familiar with induction with equal signs. We did it in high school. However I can't seem to understand out to prove by induction using the logical operator '<' (less than).

And i'm stuck on this question. All help is greatly appreciated.

So the base case is n = 7. You need to show that

3^7 < 7!

Then consider that whenever you increase n by 1, you multiply the left side by 3, and the right side by a number greater than 3. Thus the right side will continue to be greater than the left side. This is what you will write formally as the induction step. (Assume 3^n < n!. Then show that 3^(n+1) < (n+1)! is true.)
• Jun 18th 2010, 10:02 PM
atSydneyUni
OHHHH....

so we have to explicitly state that the LHS grows quicker than the RHS in the inductive step...

thank you kindly, THank you very much indeed.
• Jun 18th 2010, 10:45 PM
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Quote:

Originally Posted by atSydneyUni
OHHHH....

so we have to explicitly state that the LHS grows quicker than the RHS in the inductive step...

thank you kindly, THank you very much indeed.

You're welcome. :)

But actually you wrote that backwards; the RHS grows quicker than the LHS.