anonymous
  • anonymous
anybody good at calculus 2 ? i need help D:
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I am here
anonymous
  • anonymous
sweet! how r u at series? lol
anonymous
  • anonymous
shoot some questions

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anonymous
  • anonymous
okay well im a little bit confused about the alternating series test, how do you knw when to apply it, and do you combine it with other test as well to show that its either absolutely convergent or conditionally convergent?
anonymous
  • anonymous
i can tell you simply that if the series alternates, the sum will converge if the terms go to zero, so it is usually very easy to check
anonymous
  • anonymous
for example \[\sum\frac{1}{k}\] diverges but \[\sum\frac{(-1)^k}{k}\] converges
anonymous
  • anonymous
is that absolute or conditional convergence?
anonymous
  • anonymous
i did the ast i used the derivative i got -1/K^2 with is <0 then i used the limit and it converges towards zero
anonymous
  • anonymous
i knw that it converges by the alternating series test, am i done over all? or do i have to use another test to show that it converges absolutely or converges conditionally, cuz if u use the lct it would diverge cuz ur bn that ur comparing it with is 1/k which is a harmonic series and that converges
anonymous
  • anonymous
what was the original problem?
anonymous
  • anonymous
(-1)^k/K
anonymous
  • anonymous
oh just the one i wrote!!
anonymous
  • anonymous
ok you need no real tests for this \[\lim_{k \to \infty}\frac{1}{k}=0\] so alternating series converges
anonymous
  • anonymous
no my bad harmonic series diverges! not converge lol
anonymous
  • anonymous
however harmonic series diverges, so this clearly does not converge absolutely because it the absolute value gives the harmonic series
anonymous
  • anonymous
to recap: terms go to zero, alternating series converges abosolute value is harmonic series, diverges, so not absolutely convegent
anonymous
  • anonymous
so... do alternating series stop at the showing that it is <0 and finding the limit or do you have to continue to show the overall answer such as absolutely converges or converges conditionally
anonymous
  • anonymous
converging "conditionally" means not absolutely convergent.
anonymous
  • anonymous
i really am not sure what you are asking. all you need is that the terms to go zero
anonymous
  • anonymous
what does "it is <0 " mean?
anonymous
  • anonymous
it means that the first step in using the alternating series test is to prove that an+1is less than an either using the derivative, or an+1/an is less than 1 or an+1-an greater than zero to show that an+1 is always less that an

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