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! > ...

- anonymous

- katieb

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- watchmath

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

- anonymous

oh im sorry

- anonymous

this program is strange , i didnt see anything in it

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## More answers

- anonymous

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

- anonymous

it doesnt have to be induction

- anonymous

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

- anonymous

oh wait, that doesnt work

- anonymous

not the most elegant proof, but it might work

- watchmath

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)!

- anonymous

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

- watchmath

So you haven't read it?

- anonymous

im reading it , one sec

- watchmath

The statement \(n^2n\)
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 \(n\geq 4\).

- anonymous

ok thank you

- watchmath

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.

- anonymous

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

- anonymous

not that i know of about serious

- anonymous

because k>2 for k>=4

- anonymous

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

- watchmath

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

- watchmath

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

- watchmath

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 \(2k^2> k^2+2\geq (2k-1)+2 =2k +1\)

- anonymous

yes

- anonymous

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

- watchmath

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

- anonymous

ty

- anonymous

keeps freezing on me

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