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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)?
 8 months ago
 8 months 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)?
 8 months ago
 8 months ago

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RequiemBest ResponseYou've already chosen the best response.0
Or do i need to use a comparison test?
 8 months ago

RequiemBest ResponseYou've already chosen the best response.0
I know that those two conditions have to be met for it to be convergent...
 8 months ago

satellite73Best ResponseYou've already chosen the best response.0
is it \[\sum_{n=1}^{\infty}\frac{(1)^n}{n^3}\]
 8 months ago

satellite73Best ResponseYou've already chosen the best response.0
it not only converges, but it converges absolutely
 8 months ago

satellite73Best 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
 8 months ago

RequiemBest 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...
 8 months ago

satellite73Best 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
 8 months ago

RequiemBest ResponseYou've already chosen the best response.0
and i wasnt sure if the former was true
 8 months ago

satellite73Best ResponseYou've already chosen the best response.0
yeah only need the terms go to zero
 8 months ago

satellite73Best ResponseYou've already chosen the best response.0
\[\sum\frac{(1)^n}{n}\] for example converges, although not absolutely
 8 months ago

satellite73Best ResponseYou've already chosen the best response.0
hell, even \[\sum\frac{(1)^n}{\ln(n)}\] converges
 8 months ago

RequiemBest ResponseYou've already chosen the best response.0
its converges conditionally right?
 8 months ago

satellite73Best 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
 8 months ago

RequiemBest ResponseYou've already chosen the best response.0
ok thanks you for your help! much appreciated
 8 months ago
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