Question

prove by induction.
((n+1)!/(n+1)^(n+1)) 13 years ago

Answer 1

Have you done the base case(s) yet?

Answer 2

yeah

Answer 3

then
(n+2)!/(n+2)^(n+2) , (n+1)!/(n+1)^(n+1) goes to
((n+1)n!)/((n+2)(n+2)^n) < n!/(n+1)^n
so (n+1)!/(n+2)^(n+1) < n!/(n+1)^n < n!/n^n
so (n+1)!/(n+2)^(n+1) < n!/n^n
this is where im stuck

Answer 4

the first line shold be a < not a ,

Answer 5

Let's see then. We want to show the following.\[\frac{(n+2)!}{(n+2)^{n+2}}<\frac{(n+1)!}{(n+1)^{n+1}}\]We know that \[\frac{(n+1)!}{(n+1)^{n+1}}<\frac{n!}{n^n}\]And we also have\[\frac{(n+2)!}{(n+2)^{n+2}}=\frac{(n+2)(n+1)!}{(n+2)(n+2)^{n+1}}=\frac{(n+1)!}{(n+2)^{n+1}}\]Now, note that \[\frac{1}{(n+2)^{n+1}}<\frac{1}{(n+1)^{n+1}}\]Hence, \[\frac{(n+1)!}{(n+2)^{n+1}}<\frac{(n+1)!}{(n+1)^{n+1}}\]Which is what we wanted to show.

Answer 6

where did the inductive hypothesis come into play?

Answer 7

Now that I've looked over that, I don't think I ever used it. I don't know why you would even want to use it here. It seems like it would just complicate things.

Answer 8

yeah, seems so. We dont have to use it?

Answer 9

I've got to say, I don't even know where one could even use the inductive hypothesis. Are you sure the problem is written exactly that way?

Answer 10

well, this is to show that n!/n^n is always increasing, the book gives a hint that says it suffices to show that ((n+1)!/(n+1)^(n+1))<n!/n^n for all n in N. So I guess I dont have to use induction.

Answer 11

err decreasing*

Answer 12

In this case, I don't think you need to use induction. However, if it does say to show something for all \(n\in\mathbb{N}\), induction is usually a good bet.

Answer 13

yeah, thats why I thought it was saying use it. Great ty sir.

Answer 14

You're welcome.