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

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

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