Solve Inequality Homework: A Hint Needed

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SUMMARY

The discussion centers on proving the inequality involving double factorials, specifically the expression \(\prod_{n=1}^{1006} \frac{2n-1}{2n} < \frac{1}{\sqrt{2010}}\). The user attempted to simplify the expression by factoring out a 2 from the even terms in the denominator, leading to the form \(\frac{1}{2^{1006}}\frac{1007 \cdot 1009 \cdots 2009 \cdot 2011}{2 \cdot 4 \cdots 1004 \cdot 1006}\). The discussion highlights the relationship between double factorials and standard factorials, providing identities such as \((2k-1)! = \frac{(2k)!}{2^k k!}\) and \((2k)! = 2^k k!\) as essential tools for further simplification.

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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:

[tex] \frac{1}{2^{1006}}\frac{1007*1009*...*2009*2011}{2*4*...*1004*1006} < \frac{1}{\sqrt{2010}}[/tex]

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: [tex]\prod_{n=1}^{1006} \frac{2n-1}{2n} < \frac{1}{\sqrt{2010}}[/tex] can also be written in terms of double factorials: [tex]\frac{(2k-1)!}{(2k)!}: k=1006[/tex]

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

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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