Convergence/Divergence

[Paul McSwizzle]

I am reviewing for a math exam and am having some trouble with sequences and convergence/divergence.

One of the practice problems is:

1) Determine if the sequence {[2^(2n)]/[3^(n+2)]} from n=1 to infinity is convergent or divergent.

I am having a difficult time with this problem because of the exponents. If anyone could show me how to do this that would be great.

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.

[mickeyg13]

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I am reviewing for a math exam and am having some trouble with sequences and convergence/divergence.

One of the practice problems is:

1) Determine if the sequence {[2^(2n)]/[3^(n+2)]} from n=1 to infinity is convergent or divergent.

I am having a difficult time with this problem because of the exponents. If anyone could show me how to do this that would be great.

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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In #1, you probably want to manipulate it a bit to make it easier to work with. In other words, make it so both the numerator and denominator have the same exponent. I would start by factoring out 3^2 from the denominator. So lim ([2^(2n)]/[3^(n+2)] = lim (1/9)*[2^(2n)]/[3^n]. Now notice that power of 2n is going to be difficult to work with, so instead consider 2^(2n) as (2^2)^n=4^n. Thus you have lim (1/9)*[(4^n)/(3^n)] = lim (1/9)*(4/3)^n. Since 4/3>1, this diverges towards infinity. Of course you probably shouldn't strictly trust the results of some random message board poster that you don't know...

[mickeyg13]

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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?

[Paul McSwizzle]

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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?

[Paul McSwizzle]

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<= 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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I'm having somewhat of a tough time understanding this part of your post. Why is it now 1/2n X 1 X 1 X 1? And what does <= mean?

[mickeyg13]

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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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Yes, but "lim" was easier to type than "limit as n approaches infinity."

[mickeyg13]

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<= 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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I'm having somewhat of a tough time understanding this part of your post. Why is it now 1/2n X 1 X 1 X 1? And what does <= mean?

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<= means "less than or equal to." If we want to use the Squeeze Theorem, we need to find something that is greater than or equal to the limit in question. We can do that by replacing factors in the limit with larger factors. Each of the fractions in the limit is less than 1, so by replacing most of them with 1, we get something bigger. For something simpler, consider that 1/6 = (1/2)*(1/3) <= (1/2)*1 = 1/2. We replaced 1/3 with something larger (namely 1) in order to make the comparison.

[TomCowley]

A really simple way to do 1 is to determine the ratio of terms. If each term is successively bigger, the sequence is divergent. So take the term for n+1, (2^(2n+2))/(3^(n+3) and divide that by (2^2n)/3^(n+2). You get 4/3. So each term is 4/3 bigger than the last one, so the sequence is divergent.

[jay_shark]

2) It should be obvious that 2 approaches 0 since you have a bunch of numbers all less than 1 being multiplied out .

We use the fact that a^n = 0 as n approaches infinity for 0<=a< 1

[TomCowley]

#2 is extremely convergent, but it's possible to squeeze it to something that meets your property, but doesn't have limit 0. You can squeeze it to (2n-1)^n/(2n)^n, which is "a bunch of numbers all less than 1 being multiplied out", but converges to e^-1/2.