Question

prove induction n^2 < n! , for n>=4 ,
if it is easier n! > n^2
i didnt get far, the basis case is true
and (k+1)!= (k+1)k! > ...

Answer 1

I am sorry did you just delete your earlier question with my answer in it?

Answer 2

oh im sorry

Answer 3

this program is strange , i didnt see anything in it

Answer 4

i apologize earlier, anything you have to say is helpful :)

Answer 5

it doesnt have to be induction

Answer 6

but i have a proof of my own, the basis case is true 4! > 4^2
(k+1)! = k!(k+1) > k^2 ( 2 + 1) = k^2( 1 + 1 + 1) = k^2 + k^2 + k^2 > k^2 + 2k + 1

Answer 7

oh wait, that doesnt work

Answer 8

not the most elegant proof, but it might work

Answer 9

Ok, if you want we can use induction here we go.
So we assume that k^2 < k!
Notice that that k>1 is equivalent to 2k+1 < 3k
It follows that k>= 4 we have
2k+1 < 3k < (k)! k
Now using the assumption
k^2 < k!
we have
k^2 +(2k+1) < k! +(2k+1) + k! +(k)! k=(k+1)!
So
(k+1)^2 < (k+1)!

Answer 10

that looks, good,. one sec, what was your other proof . now im curious

Answer 11

So you haven't read it?

Answer 12

im reading it , one sec

Answer 13

The statement \(n^2<n!\) is equivalent to \(n<(n-1)!\)
Notice that for \(n\geq 4\) we have \((n-1)!\geq (n-2)(n-1)\)
So it is enough to prove \((n-2)(n-1)>n\)
The inequality is equivalent to
\(n^2-3n+2>n\)
\(n^2-4n+2>0\)
\(n(n-4)+2>0\)
which is clearly true for \(n\geq 4\).

Answer 14

ok thank you

Answer 15

Are you trying to use this for some statement about the series? if it is the case maybe it will be easier to prove the statement for some bigger n (not 4) to make it more obvious.

Answer 16

thats clever, can you look over my proof
the basis case is true 4! > 4^2 (k+1)! = k!(k+1) > k^2 ( 2 + 1) = k^2( 1 + 1 + 1) = k^2 + k^2 + k^2 > k^2 + 2k + 1

Answer 17

not that i know of about serious

Answer 18

because k>2 for k>=4

Answer 19

oh but then i need to show k^2 > 2k ?

Answer 20

correct, that is what I was going to say :)

Answer 21

well actually you need 2k^2 > 2k+1

Answer 22

Notice that
\((k-1)^2 \geq 0\)
So \(k^2-2k+1 \geq 0\)
So \(k^2\geq 2k-1\)
For \(k\geq 4, k^2 > 2\)
So \(2k^2> k^2+2\geq (2k-1)+2 =2k +1\)

Answer 23

yes

Answer 24

can i ask a quick question. im helping a student who thinks that you can form a solid of revolution , bounded by y=x^2, y = x+2, and revolve it about the y axis

Answer 25

Yes I think you can but there is some overlap between the solid form by the region on the right of the y-axis and the solid form by the region on the left side of the y-axis

Answer 26

ty

Answer 27

keeps freezing on me