Re: Convergence/Divergence
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The second problem is:
2) Use the squeeze theorem to show that the sequence {[1x3x5x7...x(2n-3)x(2n-1)]/[(2n)^n] from n=1 to infinity is convergent to 0.
I really don't understand how to use the squeeze theorem here. Again, any help would be great.
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For #2, look at the numerator. You are multiplying all odd positive integers from 1 to 2n-1. Ask yourself how many such integers there are. Hopefully you realize there are n such odd integers from 1 to 2n-1 (and n-1 even ones). So in both the numerator and denominator, we are multiplying n factors (the denominator has n factors of 2n).
Now let's break up the fraction a bit. We have n factors in the numerator and n factors in the denominator, so let's pair them off and instead multiply together n fractions. Thus (1*3*5*...*(2n-3)*(2n-1))/(2n)^n becomes (1/(2n))*(3/(2n))*(5/(2n))*...*((2n-3)/(2n))*((2n-1)/(2n)). Now observe that all of these factors are proper fractions, meaning that they are all less than 1. Thus multiplying together more of the factors gives us something smaller than if we only look at the first factor (namely (1/(2n)). Therefore lim (1*3*5*...*(2n-3)*(2n-1))/(2n)^n = lim (1/(2n))*(3/(2n))*(5/(2n))*...*((2n-3)/(2n))*((2n-1)/(2n)) <= lim (1/(2n)*1*1*...*1*1 = 0. Of course 0 <= lim (1*3*5*...*(2n-3)*(2n-1))/(2n)^n, so by the Squeeze Theorem lim (1*3*5*...*(2n-3)*(2n-1))/(2n)^n = 0. Make sense?
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You mean limit as n approaches infinity?
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