1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Alternating Series and P-series convergence

  1. Jul 28, 2008 #1
    Alternating Series and P-series "convergence"

    I couldn't resist trying out a pun. Anyway, onto the question:

    1. The problem statement, all variables and given/known data
    Test the series for convergence/divergence:
    [tex]\sum^{\infty}_{n=1}\frac{-1^{n-1}}{\sqrt{n}}[/tex]


    2. Relevant equations
    Alternating Series Test and possibly p-series test...


    3. The attempt at a solution
    The expression for an alternating series goes as a(n) = (-1)^(n-1) * b(n). Having said that, it's obvious that b(n) is

    [tex]\frac{1}{\sqrt{n}}[/tex]

    which can be re-written as [tex]\frac{1}{n^\frac{1}{2}}[/tex]. But by the rules of the p-series, since 0.5 is obviously lesser than 1, I came to the conclusion that the series diverges. Instead, the answer states that the series does indeed converge. Can anyone help shed some light in this?
     
  2. jcsd
  3. Jul 28, 2008 #2

    rock.freak667

    User Avatar
    Homework Helper

    Re: Alternating Series and P-series "convergence"

    I think for the p-series test, the numerator must be '1' only.
     
  4. Jul 28, 2008 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Alternating Series and P-series "convergence"

    Forget the p-series test. Concentrate on the alternating series test. b(n) is decreasing towards 0. Sorry, I don't get the 'pun'. Am I being thick?
     
  5. Jul 28, 2008 #4
    Re: Alternating Series and P-series "convergence"

    I don't think you are since I don't get it either.
     
  6. Jul 28, 2008 #5

    rock.freak667

    User Avatar
    Homework Helper

    Re: Alternating Series and P-series "convergence"

    Well usually for the p-series test, the only examples with fractions I've ever seen is where the function is in the form

    [tex]\frac{1}{f(n)}[/tex]
     
  7. Jul 28, 2008 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Alternating Series and P-series "convergence"

    Sure, in fact, it applies only to series of the form 1/n^p or n^(p). I wasn't saying you were wrong. It's useful here if you want to discuss absolute convergence. It is useless for the given function.
     
  8. Jul 29, 2008 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Alternating Series and P-series "convergence"

    The "p-series" only applies to positive series. That's why Dick says it is useful to discuss absolute convergence. If your series has both positive and negative terms then it may converge "conditionally". It will converge absolutely only if the series of absolute values converges.

    By the "alternating series test", the series
    [tex]\sum \frac{(-1)^n}{\sqrt{n}}[/tex]
    converges.

    By the "p- series test"
    [tex]\sum \left|\frac{(-1)^n}{\sqrt{n}}\right|= \sum \frac{1}{\sqrt{n}}= \sum \frac{1}{n^{1/2}}[/tex]
    does NOT converge and so the original series converges conditionally, not absolutely.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Alternating Series and P-series convergence
Loading...