Test series for absolute convergence
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Test series for absolute convergence

[From: ] [author: ] [Date: 11-05-30] [Hit: ]
Using ratio test, I get L = 1, which is inconclusive.It is not a harmonic series, nor a geometric series, nor a p-series.......
1 - (1*3)/3! + (1*3*5)/5! - (1*3*5*7)/7! + .... + (-1)^(n-1)[1*3*5...(2n-1)]/(2n-1)!

It appears it would be absolutely convergent, because all of the odd factors would cancel out with the factorial in the denominator, leaving only even factors in the denominator, however I need to mathematically show that it is absolutely convergent.

Using ratio test, I get L = 1, which is inconclusive. It is not a harmonic series, nor a geometric series, nor a p-series. I wouldn't think integral test or comparison test would work for this type of series. So what do I do? I know it is conditionally convergent for sure, because without the (-1), the limit is 0, and sequence is decreasing as n increases. I'm fresh out of ideas.

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Just glancing at your work ...

a(n) = (-1)^(n-1)[1*3*5...(2n-1)]/(2n-1)!

a(n+1) = (-1)^n[1*3*5...(2n+1)]/(2n+1)!

|a(n+1)/a(n)| = (2n+1)/{ (2n+1)*(2n) } = 1/(2n)

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Let

a_n = [1*3*...*(2n-1)]/[(2n-1)!]

Since the numerator is the product of all the odd numbers up to 2n-1, and the denominator is the product of all numbers up to 2n-1, you have by some cancellation that

a_n = 1/(2 * 4 * ... * (2n-2))

at least whenever n > 1 (so that the product indicated by the ... in the denominator actually has terms in it). Whenever n > 2 the product in the denominator has at least the terms 2n-2 and 2n-4 in it, and perhaps other things as well. Since removing these other things from the denominator can only make the fraction bigger, we conclude

a_n <= 1/((2n-2)(2n-4)) for all n > 2.

The series sum 1/((2n-2)(2n-4) is easily seen to be absolutely convergent (e.g. by limit comparison with 1/n^2), so by the comparison test, sum a_n is convergent too. This means your series is absolutely convergent.

For another way to do this problem, you could actually use the limit comparison test with any 1/n^p with p > 1 (by which I mean: for any p > 1, you can show that the limit of a_n/(1/n^p) is 0, so by the limit comparison test, since sum 1/n^p is convergent, so is sum a_n). But for some reason I find it easier to think in terms of the comparison test.
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