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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)?

Calculus1
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Or do i need to use a comparison test?
I know that those two conditions have to be met for it to be convergent...
is it \[\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}\]

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Other answers:

yes thats it
it not only converges, but it converges absolutely
all you need for the alternating series to converge is that the terms go to zero, which thsee certainly do
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...
but \[\sum_{n=1}^{\infty}\frac{1}{n^3}\] also converges, so this series is not just convergent, it is absolutely convergent as well
and i wasnt sure if the former was true
yeah only need the terms go to zero
ok, thanks satellite
\[\sum\frac{(-1)^n}{n}\] for example converges, although not absolutely
hell, even \[\sum\frac{(-1)^n}{\ln(n)}\] converges
its converges conditionally right?
the last two i wrote are only conditionally convergent, the one you posted is absolutely convergent, a stronger condition
ok thanks you for your help! much appreciated
yw

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