PoShan L. answered • 01/29/15

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Step 1: Shows inequality holds for n = 1, I will leave that to you to show.

Step 2: Then you want to show that IF the inequality holds for n, then it also holds for n + 1.

Assume the inequality holds for n, then you

**have**the following:2!*...*(2n)! >= ((n+1)!)

^{n}------(eq 1)Now you

**need**to show that the inequality also holds for n + 1, so you**need**to show2!*...*(2n)!*(2(n+1))! >= (((n+1)+1)!)

^{n+1}------(eq 2)Now, let's multiply (2(n+1))! to both sides of (eq 1), here we assume that n are positive integers, therefore the sign of the inequality does not change, and we have,

2!*...*(2n)!*(2(n+1))! >= ((n+1)!)

^{n}*(2(n+1))! ------(eq 3)Compare (eq 3) to (eq 2), notice that the left hand side is what you what, and we need to make the right hand side of (eq 3) the same as (eq 2).

R.H.S of (eq 3) ((n+1)!)

^{n}*(2(n+1))! =((n+1)!)^{n}*(2(n+1))*...*(n+3)*((n+1)+1)! (Similar to writing 5! = 5*4*3!)Now realizing all terms larger than ((n+1)+1) in (2(n+1))! is larger than ((n+1)+1), and there are exactly n those terms, you can replace each of those larger term by ((n+1)+1) and get

((n+1)!)

^{n}*(2(n+1))*...*(n+3)*((n+1)+1)! > ((n+1)!)^{n}*((n+1)+1)^{n }*((n+1)+1)!Now combine each (n+1)! with ((n+1)+1) to get ((n+1)+1)!,

((n+1)!)

^{n}*((n+1)+1)^{n}*((n+1)+1)! = (((n+1)+1))!)^{n}*((n+1)+1)!Finally, combing the power

(((n+1)+1))!)

^{n}*((n+1)+1)!^{1}= (((n+1)+1))!)^{n+1}If you still remember, we start off with the R.H.S of (eq 3), so we proved

((n+1)!)

^{n}*(2(n+1))! >(((n+1)+1))!)^{n+1}but we are not done. By transitive property of inequality, we combine

2!*...*(2n)!*(2(n+1))! >= ((n+1)!)

^{n}*(2(n+1))! ------(eq 3) ((n+1)!)

^{n}*(2(n+1))! >(((n+1)+1))!)^{n+1}to get

2!*...*(2n)!*(2(n+1))! >= (((n+1)+1))!)

^{n+1 }So finally finally, we prove that the inequality holds for n + 1

Yuri Y.

Thank you so much. I very appreciate this.

I find the answer very helpful but I'm still confused with some parts

(2(n+1))! =(2(n+1))*...*(n+3)*((n+1)+1)!

Why it's not (2(n+1))! = (2n+2) * (2n+1) * 2n * (2n-1) *.....

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01/29/15

PoShan L.

I am glad you find it helpful and I am very glad that you wonder about that. Let me clarify.

First of all, what you wrote is correct, HOWEVER, it is not useful, it doesn't go anywhere. Now, why did I write it the way I wrote, the answer to that is, 1) it is correct, 2) it works.

Let me explain 1), I am assuming you agree that (2(n+1)) is the same thing as (2n + 2), so we are not disagreeing on the first part, now I want to point your attention to the
! in ((n+1)+1)!, in case you missed it, it is important that there is !. If the ! is missing, it is wrong. So what I did was kind of like writing

5! = 5*4*3!

do you agree that it is the same? If you are not sure, look at this,

R.H.S = 5*4*3! = 5*4*3*2*1

Now if you agree, go back to look at

(2(n+1))! =(2(n+1))*...*(n+3)*((n+1)+1)!

do you see it this time? If you still don't see it, go ahead and make up a number for n, and expand out all the terms in the !, then you will see that they are indeed equal.

Now, the motivation of doing it my way is because I see that the R.H.S still has a ! into it, so I know that I need to repackage the terms into a !.

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01/29/15

PoShan L.

01/29/15