Solve Inequality Homework: A Hint Needed

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The discussion revolves around proving an inequality involving factorials and double factorials. The user has simplified the expression by factoring out a 2 from the even terms in the denominator, leading to a comparison with a value less than 1 over the square root of 2010. They seek guidance on how to proceed from their current expression, which involves products of odd integers and even integers. The conversation also references identities related to factorials that may assist in the proof. The focus remains on finding a hint to advance the solution of the inequality.
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Homework Statement


To prove the inequality (attached)


Homework Equations


The Attempt at a Solution



I tried factoring out a 2 from each of the even terms in the denominator. This allowed me to cancel out all the terms (odd) on the numerator up to 1005.

Leaves me with:

<br /> \frac{1}{2^{1006}}\frac{1007*1009*...*2009*2011}{2*4*...*1004*1006} &lt; \frac{1}{\sqrt{2010}}<br />

I don't know how to continue after this point. Can someone please give me a hint? Thanks.
 

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um, that would be: \prod_{n=1}^{1006} \frac{2n-1}{2n} &lt; \frac{1}{\sqrt{2010}} can also be written in terms of double factorials: \frac{(2k-1)!}{(2k)!}: k=1006

Then you have things like: (2k-1)!=\frac{(2k)!}{2^k k!} \qquad \text{and} \qquad (2k)!=2^k k!

http://en.wikipedia.org/wiki/Factorial#Double_factorial
... the identities will be standard proofs you can look up.

notice that 2010 = 2(k-1)
 
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