Need help with an Induction Proof!

  • #1

Homework Statement


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Prove the following:
(1x3x5x...x(2n-1))/(2x4x6x...x(2n)) <= 1/(2n+1)^.5


Homework Equations



none

The Attempt at a Solution


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1) Following the general steps of proof by induction, I first proved that the statement is true for the case n=1.

2) Next, I substituted n+1 for n into the equation.

3) Using algebra, I have hammered away at the problem from every angle that I can think of in hopes to end up with the same formula as in the original case. Is this the correct way to go about solving this problem? My best attempt thus far is in the attached file. My work begins with my substitution of n+1 for n.



 

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Answers and Replies

  • #2
RUber
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##1/2 \leq 1/\sqrt3##
Assume that for ##n, \, \prod_{i=1}^n \frac{2n-1}{2n} \leq \left( \frac{1}{2n+1} \right) ^{1/2}##
Then show that it holds true for ##n+1##
The goal is to use the fact that the relation is true for n so you only have to show that the addition of one term doesn't change the fact that the left side is less than the right side.
Think of it as the change to the left divided by the change to the right must be less than or equal to 1 for the induction to work.
 
  • #3
##1/2 \leq 1/\sqrt3##
Assume that for ##n, \, \prod_{i=1}^n \frac{2n-1}{2n} \leq \left( \frac{1}{2n+1} \right) ^{1/2}##
Then show that it holds true for ##n+1##
The goal is to use the fact that the relation is true for n so you only have to show that the addition of one term doesn't change the fact that the left side is less than the right side.
Think of it as the change to the left divided by the change to the right must be less than or equal to 1 for the induction to work.

I took this approach in my attached work, but still couldn't get the induction to work. Do I have to use some kind of squeeze theorem approach you think?
 
  • #4
RUber
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Think about the net change from the ##n^{th} ## evaluation.
On the left side, you are multplying ##x_n## by ##\frac{2n+1}{2n+2}##.
The right side, you could look at as multiplying by ##\sqrt{\frac{2n+1}{2n+3}}##.

Call the nth evaluation of the left side ##L_n##, and the right side ##R_n## or whatever you want.
##L_{n+1}=L_n \frac{2n+1}{2n+2}##
##R_{n+1}=R_n \sqrt{\frac{2n+1}{2n+3}}##.

You need the change on the left to be less than or equal to the change on the right for the induction to work.

You know ##\frac{L_n}{R_n} \leq 1##
You need to show ##\frac{L_{n+1}}{R_{n+1}}\leq 1##
 
  • #5
RUber
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There may be times when flipping the fractions is helpful. Remember ## 1/n < a/b \implies n/1 > b/a . ##
 
  • #6
Thanks for the help! I'll use this approach and see if I can figure it out.
 

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