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How to apply comparison test?

  1. Apr 15, 2013 #1
    1. The problem statement, all variables and given/known data

    Is the series convergent or divergent? [tex]\sum_{n=0}^{\infty}{\frac{1}{\sqrt{n+1}}}[/tex]

    2. Relevant equations

    I can use any test but wolfram alpha says that it is divergent by comparison test.

    3. The attempt at a solution

    How do I apply comparison test?

    I can compare it to: [tex]\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ \sqrt { n } } }[/tex] but the second series is greater than the series in the question and the second series is divegent. :(
     
  2. jcsd
  3. Apr 15, 2013 #2

    LCKurtz

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    Have you had the "generalized comparison test"? That's what you want.
     
  4. Apr 15, 2013 #3
    You mean the limit comparison test? Yes, I've learned both the comparison test and the limit comparison test.

    Currently trying to use the second series for limit comparison test.
     
  5. Apr 15, 2013 #4

    LCKurtz

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    Yes. That should work.
     
  6. Apr 15, 2013 #5
    Wait. I found an easier method. Just let k=n+1.
     
  7. Apr 15, 2013 #6

    LCKurtz

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    Yes, that's pretty slick. Still, the limit comparison test is your friend and you want to be adept at recognizing when and how to use it. Your idea won't work on lots of similar problems that are just a little bit more complicated.
     
  8. Apr 15, 2013 #7
    I just realized that my second limit does not make sense since when n=0, you get 1/0.

    What do I use for limit comparison test? :(
     
  9. Apr 15, 2013 #8

    LCKurtz

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    That doesn't matter. It only matters what happens at the tail end of the series. Just leave off the first term; it won't affect convergence.

    Use your original ##\frac 1{\sqrt n}## for the limit comparison test.
     
  10. Apr 15, 2013 #9
    Haha. Got it. Thank you!
     
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