anonymous
  • anonymous
why cant i apply the alternating series test to this series sigma n=1 to infinity (-1)^n-1 sin n
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Says here, to use alternating it should be of the for -1^(n+1) or -1^n
anonymous
  • anonymous
is there any other explanation besides that
anonymous
  • anonymous
No. The point is there are many test and any one of them may be inconclusive and you may have to use another one in conjunction with it. Flexibility is needed on your part.

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anonymous
  • anonymous
so if there was a minus sign n-1 then it would be alternating?
anonymous
  • anonymous
I am not sure what you are referring to.
anonymous
  • anonymous
"Says here, to use alternating it should be of the for -1^(n+1) or -1^n 10 minutes ago "
anonymous
  • anonymous
Yeah. Yours is not like that. Yours has n-1 in the exponent.
anonymous
  • anonymous
so those are the only two main condition that cause it to have alternative signs
anonymous
  • anonymous
I think its -1 in front that makes it alternating. Yours is alternating but they don't like the n-1 in the exponent
anonymous
  • anonymous
when i plugged in values i gott two negatives and then two positive
anonymous
  • anonymous
Yep, that's what they mean by alternating
anonymous
  • anonymous
now im kinda of confused then why is it that in this problem we can apply the test if thats what happens
anonymous
  • anonymous
OK, with the power vested in me, I hereby give you permission to use the alternating series test.
anonymous
  • anonymous
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