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.
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)
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.
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.