The discussion focuses on finding an expression E(n) for the ratio of n!/(1*3*...*(2*n+1)) as n approaches infinity. Participants highlight the use of the relationship 1*3*5*...*(2n+1) = (2n+1)!/[2*4*6*...*2n] to simplify the denominator. The goal is to derive an analog of Stirling's formula for the product of odd integers, specifically ∏(2*k+1). This approach leads to a solution that satisfies the original problem statement.
PREREQUISITESMathematicians, students studying combinatorics, and anyone interested in asymptotic analysis and factorial properties.
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
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