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Expert: - 9/17/2010


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

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

"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
There are three things to note about this problem which show that the left side is greater.

1 – there are the same number of terms multiplied together on both sides of the equation.  That is, they both have n terms.

2 – The curve the points are taken from is concave up.  That is, the higher the curve gets, the faster they are increasing.

3 – There is always a difference of 2 between the terms on the left.  This makes the n values as far above the average as the terms below the average.

For example, if there were 4 terms,  the left side would have 2!, 4!, 6!, and 8!.  The right side would be 5!^4.  The first and last terms on the left are 2 and 8, and these both differ from 5 by 3.  The middle two terms are 4 and 6, and these both differ from 5 by 1.
Since the curve is concave up, it is known that 1!*5! > 3^2 and 2!*4! > 3!^2.  Multiplying these together gives so 1!*2!*3!*4! > 3!^4.

If there is an even number of terms, this can be done.
If there is an odd number of terms, do the same, ignoring the middle term.

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