Help Needed: Understanding An=1x3x5... (2n-1)/(2n)^n Convergence

[anderma8]

I'm trying to figure out the following:

An=1x3x5... (2n-1)/(2n)^n and I'm trying to determine if it converges or diverges and if it converges, what the limit is. The answer is 1/2n x 3/2n x 5/2n... (2n-1)/2n and it converges, but I don't understand what they did or how they got to the answer. Any help?

 

[HallsofIvy]

I'm trying to figure out the following:

An={1x3x5... (2n-1)}/(2n)^n and I'm trying to determine if it converges or diverges and if it converges, what the limit is. The answer is 1/2n x 3/2n x 5/2n... (2n-1)/2n and it converges, but I don't understand what they did or how they got to the answer. Any help?

Your "answer" does not give the limit. What they have done, very simply, is take that (2n)^n in the denominator and distribute it to each of the factors in the numerator. For example, if n= 3 you have {1*3*5}/(6^3)= (1/6)(3/6)(5/6). Since each factor is less than 1, and multiplying any positive number by a number less than one gives a smaller product, multiplying by more and more such fractions makes the product smaller and smaller- the sequence clearly converges to 0.

 

[tacman]

Firstly, it is important to note that the expression given can be rewritten as An = (2n-1)^1/(2n)^n. This may make it easier to understand the steps taken to determine convergence.

To determine convergence, we can use the ratio test, which states that if the limit of the absolute value of the ratio of consecutive terms is less than 1, then the series converges. Let's apply this to our expression:

lim┬(n→∞)⁡〖|((2(n+1)-1)^1/(2(n+1))^n )/((2n-1)^1/(2n)^n )|〗

= lim┬(n→∞)⁡〖|((2n+1)^1/(2n+2)^n )/((2n-1)^1/(2n)^n )|〗

= lim┬(n→∞)⁡〖|((2n+1)/(2n-1))^1/((2n+2)/(2n))^n |〗

= lim┬(n→∞)⁡〖|((2n+1)/(2n-1))^1/((2n+2)/(2n))^1 |〗

= lim┬(n→∞)⁡〖|(2n+1)/(2n-1)|〗 (since n→∞, the exponent n becomes 1)

= 1 (since the highest power of n in the numerator and denominator is 1)

Since this limit is less than 1, we can conclude that the series converges.

Now, to find the limit, we can use the fact that the ratio of consecutive terms approaches a constant value as n→∞. In this case, the constant value is 1. Therefore, we can write:

lim┬(n→∞)⁡〖(2n-1)^1/(2n)^n 〗= lim┬(n→∞)⁡〖1/(2n)^n 〗

= 1/lim┬(n→∞)⁡〖(2n)^n 〗 (since the limit is in the denominator, we can flip the fraction)

= 1/(∞)