What is n/(1*3* *(2*n+1)) for large n?

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Homework Statement


What is n!/(1*3*...*(2*n+1)) for large n ? (I need an expression E(n) such that the ratio of E(n) by the quantity (n!/(1*3*...*(2*n+1)) goes to 1 as n goes to ∞.

Homework Equations



Does anybody know of an analog of Stirling's formula for ∏(2*k+1)?

The Attempt at a Solution



 
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htg said:

Homework Statement


What is n!/(1*3*...*(2*n+1)) for large n ? (I need an expression E(n) such that the ratio of E(n) by the quantity (n!/(1*3*...*(2*n+1)) goes to 1 as n goes to ∞.

Homework Equations



Does anybody know of an analog of Stirling's formula for ∏(2*k+1)?

The Attempt at a Solution


Use the fact that 1*3*5* ... *(2n+1) = (2n+1)!/[2*4*6* ... *2n], and re-write the denominator here.

RGV
 
Ray Vickson said:
Use the fact that 1*3*5* ... *(2n+1) = (2n+1)!/[2*4*6* ... *2n], and re-write the denominator here.

RGV

Thanks, it solves my problem.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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