Question

Prove that : \((n!)^2 > n^n\) for n>2

Answer 1

you would have to use induction, correct?

Answer 2

I don't think so. May be, induction may work here but I suppose, there is some other better method. Not sure!

Answer 3

I see a problem with the statement. It is only true for n>2.

Answer 4

Forgot to add that @Paynesdad .

Answer 5

you could prove it by induction, that is the only way how i could see proving this inequality

http://math.stackexchange.com/questions/170173/proving-inequalities-using-induction

Answer 6

Would expressing the product as a sum help? Meaning, take the log of both sides, then you have the sum of \(n\) terms on the left and one term on the right. Something like
\[2\sum_{k=0}^n\ln(n-k)>n\ln n\]

Answer 7

What about using wavy curve method?

Answer 8

Sorry, should be\[2\sum_{k=0}^{n-1}\ln(n-k)>n\ln n\]
Not sure if it helps...

Answer 9

Siths, yep, I am getting that but also not sure how to continue from that ...

Answer 10

If anything, it might help in making a simpler induction proof. Starting with the base case \(n=3\), you have
\[2\ln3+2\ln2+2\ln1>3\ln3\\
2\ln2>\ln3\\
\ln4>\ln3\]

Answer 11

If you can find a way of expressing this inequality for the \(n\)th term, I think you'll be in luck.

Answer 12

Thinking...

Answer 13

hmm I think I have a solution have a look
first expand (n!)^2 as = n^2*(n-1)*[(n-2)*1]*[(n-2)*2]............[1*(n-2)]
also write n^2 as n*n*n......n (n terms)

Answer 14

hmm, since none ofthe parts equal zero\[(n!)^2 > n^n\]
\[\frac{(n!)^2}{n^n} > 1\]
\[\frac{n^2[(n-1)!]^2}{n^n} > 1\]
\[n^{2-n}[(n-1)!]^2 > 1\]
maybe?

Answer 15

now first cancel n^2 from both the sides.....so on right we have n^n-2

Now divide n with each of the product pair i.e (n-2)*1/n.........(n-k-1)*(k-1)/n....1*(n-2)/n

Answer 16

Nice @amistre64 and @ankit042 , I will respond about the solns you both have given, for now, i have to go for dinner :)

Answer 17

Wait...rethinking...

Answer 18

how about l'hopital?

Answer 19

\[n^{2-n}[(n-1)!]^2 > 1\]
\[[(n-1)!]^2 > \frac{1}{n^{2-n}}\]
im sure this would be self evident :)

Answer 20

It can't be that simple^
D:

Answer 21

so on left we will have (n-1)[there will be n-2 pairs of type (n-k-1)k/n where 1<=k<=n-2]
now the minimum value for a product pair will be when k =n/2 [THINK]
so use this value for all k.
We will get (n-1)(n-2)/2 >1 which is true for all n >2

Answer 22

take the log, get
\[2\ln(n!)>n\ln(n)\] or \[\frac{n\ln(n)}{2\ln(n!)}<1\]
yeah that is silly, that is a limit
true one though

Answer 23

mines got a little flaw .. 2-n doesnt stay a fraction does it

Answer 24

Guys this might not be very clear but have a look

Answer 25

ahhh .... never try to solve math problems when you have to use the bathroom :)

Answer 26

Okay, a new idea...Let
\[\Large a_n = n^n\]\[\Large b_n = (n!)^2\]

Then
\[\Large \frac{a_n}{b_n}\]
(after the dust settles)
\[\Large \frac{a_n}{b_n}=\left( \frac{n+1}{n}\right)^n \cdot \frac1{n+1} \]

Which, in terms of x...
\[\Large \left( \frac{x+1}{x}\right)^x \cdot \frac1{x+1} \]
Is a decreasing function from x \(\ge\) 2

Answer 27

So, this
\[\Large \frac{a_n}{b_n}\] is decreasing, as n goes to infinity, with n starting at 3.
In fact, at n = 3, you have
\[\Large \frac{a_n}{b_n}=\frac{27}{36}= \frac34 < 1\]
And it's only going to get smaller as n gets larger, so, effectively, for all n \(\ge\) 3

\[\Large \frac{a_n}{b_n}<1\]\[\Large a_n <b_n\]\[\huge \color{blue}{n^n < (n!)^2}\]

Answer 28

Nice one @terenzreignz