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- anonymous

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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- anonymous

- chestercat

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- anonymous

Or do i need to use a comparison test?

- anonymous

I know that those two conditions have to be met for it to be convergent...

- anonymous

is it
\[\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}\]

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- anonymous

yes thats it

- anonymous

it not only converges, but it converges absolutely

- anonymous

all you need for the alternating series to converge is that the terms go to zero, which thsee certainly do

- anonymous

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...

- anonymous

but \[\sum_{n=1}^{\infty}\frac{1}{n^3}\] also converges, so this series is not just convergent, it is absolutely convergent as well

- anonymous

and i wasnt sure if the former was true

- anonymous

yeah only need the terms go to zero

- anonymous

ok, thanks satellite

- anonymous

\[\sum\frac{(-1)^n}{n}\] for example converges, although not absolutely

- anonymous

hell, even
\[\sum\frac{(-1)^n}{\ln(n)}\] converges

- anonymous

its converges conditionally right?

- anonymous

the last two i wrote are only conditionally convergent, the one you posted is absolutely convergent, a stronger condition

- anonymous

ok thanks you for your help! much appreciated

- anonymous

yw

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