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Requiem
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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
 one year ago
Requiem Group Title
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
 one year ago

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

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

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

Requiem Group TitleBest ResponseYou've already chosen the best response.0
yes thats it
 one year ago

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

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

Requiem Group TitleBest ResponseYou've already chosen the best response.0
oh 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...
 one year ago

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

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

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

Requiem Group TitleBest ResponseYou've already chosen the best response.0
ok, thanks satellite
 one year ago

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

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

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

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

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