Expert: Ahmed Salami - 9/17/2010

Question
This is a hard question, and it needs to be completed using the principle of mathematical induction, That is, prove


(2!)(4!)(6!)...(2n)! >= [(n+1)!]^n


My Note**"Taking a look at the question, there is no restriction on any n intergers, nor any limits defined on n->. I have tried the idea of containg cases, such that define a particular series. I have had troulbe. Perhaps, if you can implement limits somehow, or try using cases, I ask that you do. Any help would be great. My confusion lies after the factorials are split;

Answer
Hi William,
The idea of the principle of mathematical induction is to prove that;
i) The expression is true for n = 1
ii) And that it would be true for n+1 whenever it is true for n
So, for n = 1, we have
2! >= [(1+1)!]^1
2! >= 2!
which is true.
If its true for n, then
(2!)(4!)(6!)...(2n)! >= [(n+1)!]^n
We need to prove that its also true for n+1 as a consequence of it being true for n. If its true for n+1, we would have
(2!)(4!)(6!)...(2n)!.[2(n+1)]! >= [(n+2)!]^(n+1)

There should be a number of ways to consider this depending on what comes to mind. For instance, we could start with the use of the expression
(n+2)! > 2^n for all positive n
You might also need to see the proof of this which i provide later, but lets for the moment take it as true.
This means that
a) [(n+1)!]^n > [2^(n-1)]^n
[(n+1)!]^n > 2^(nē-n)

b) [(2n+2)]! > 2^(2n)

All we need do now is prove that
(2!)(4!)(6!)...(2n)!.[2(n+1)]! > [2^n]^(n+1)
i.e (2!)(4!)(6!)...(2n)!.[2(n+1)]! > 2^(nē+n)
whenever
(2!)(4!)(6!)...(2n)! >= [(n+1)!]^n
i.e (2!)(4!)(6!)...(2n)! > 2^(nē-n)
From the last line,
(2!)(4!)(6!)...(2n)!.[2(n+1)]! > 2^(nē-n) . 2^(2n)
(2!)(4!)(6!)...(2n)!.[2(n+1)]! > 2^(nē+n)
and the proof is complete.


For the proof of (n+2)! > 2^n
Its true for n = 1 since 3! > 2
If its true for n, then
(n+2)!.(n+3) > (2^n).(n+3)
but n+3 > 2 for all positive n, and so
(n+3)! > (2^n).2
(n+3)! > 2^(n+1)
which is the statement for n+1 and hence the proof.

I do hope this satisfies you and i wish you all the best. And you can always get back to me.

Regards