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 one year ago
Determine convergence or divergence.
(1)^n/(n^3)
I know that I want to use the alternating series test. I used (bsubn) as 1/n^3 and (bsubn+1) as 1/(n+1)^3. I know that (bsubn) as n > infinity equals 0, but is (bsubn+1) less than or equal to (bsubn)?
 one year ago
Determine convergence or divergence. (1)^n/(n^3) I know that I want to use the alternating series test. I used (bsubn) as 1/n^3 and (bsubn+1) as 1/(n+1)^3. I know that (bsubn) as n > infinity equals 0, but is (bsubn+1) less than or equal to (bsubn)?

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Requiem
 one year ago
Best ResponseYou've already chosen the best response.0Or do i need to use a comparison test?

Requiem
 one year ago
Best ResponseYou've already chosen the best response.0I know that those two conditions have to be met for it to be convergent...

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0is it \[\sum_{n=1}^{\infty}\frac{(1)^n}{n^3}\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0it not only converges, but it converges absolutely

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0all you need for the alternating series to converge is that the terms go to zero, which thsee certainly do

Requiem
 one year ago
Best ResponseYou've already chosen the best response.0oh ok..in my book it says that (bsubn+1) has to be less than or equal to (bsubn) but also that (bsubn) has to equal 0 when the limit is taken...

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0but \[\sum_{n=1}^{\infty}\frac{1}{n^3}\] also converges, so this series is not just convergent, it is absolutely convergent as well

Requiem
 one year ago
Best ResponseYou've already chosen the best response.0and i wasnt sure if the former was true

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0yeah only need the terms go to zero

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum\frac{(1)^n}{n}\] for example converges, although not absolutely

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0hell, even \[\sum\frac{(1)^n}{\ln(n)}\] converges

Requiem
 one year ago
Best ResponseYou've already chosen the best response.0its converges conditionally right?

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0the last two i wrote are only conditionally convergent, the one you posted is absolutely convergent, a stronger condition

Requiem
 one year ago
Best ResponseYou've already chosen the best response.0ok thanks you for your help! much appreciated
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