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Convergence Test (Comparison) Questions

  1. May 18, 2013 #1
    Hello everyone,

    I need some help on doing convergence tests (comparisons I believe) on some Ʃ sums.

    I have three, they are:

    1. Ʃ [ln(n)/n^2] from n=1 to ∞.

    I tried the integral test but was solved to be invalid (that is, cannot divide by infinity). Therefore I believe it to be a comparison test but don't know what to compare it to.

    2. Ʃ [(3n+2)/(n^3+1)] from n=0 to ∞

    I solved the whole problem starting at n=1 and stating that Ʃ1/n^2 and Ʃ1/n^3 were larger and convergent, therefore the sum is convergent. I realised the n=0 part later and now cannot solve it cause you cannot divide by 0; that is, 1/0^2 or 1/0^3. I tried an index shift yet it still is invalid.

    3. Ʃ [(2n+1)/n^2] from n=1 to ∞

    I don't know what to compare this to. I need a value smaller than 1/n^2 which diverges (as sum diverges) but 1/n^3 converges. I don't know what to do.

    Help would be much appreciated.

    Thank you
     
  2. jcsd
  3. May 18, 2013 #2

    mfb

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    Why would you want to "divide by infinity"?
    A comparison can be useful - you'll need something "between" 1/n (-> sum is not convergent) and 1/n^2 (->sum is convergent, but it is not larger than your expression)

    For convergence, finite numbers of initial summands are irrelevant. It is sufficient if your comparison works for n>2, n>1 billion or any other number.

    The sum over 1/n^2 converges. You can split this into two series, both of them are easier to evaluate afterwards.
     
    Last edited: May 19, 2013
  4. May 18, 2013 #3
    How should I go about determining which 'in between' fraction is convergent? Do I use trial and error, and then solve it geometrically?

    Okay thanks for that so when I solve, I can just say that all values are positive for n≥1 and leave it to that. Then I solve without considering n=0.

    I tried something like that but kind of got stuck.

    a_n = (2n+1)/n^2 = 2n/n^2 + 1/n^2......1/n^2 is already convergent so don't need to worry about it.

    2n/n^2=2/n......n > x? n is larger than what convergent value so that when I invert it, 1/n is smaller than 1/x --the convergent value.

    Then do I say that 2/n < 1/x = convergent + 1/n^2 convergent = Ʃ sum is convergent by Comparison Test.
     
    Last edited: May 18, 2013
  5. May 19, 2013 #4

    micromass

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    The harmonic series ##\sum \frac{1}{n}## is not convergent.
     
  6. May 19, 2013 #5

    mfb

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    What could be interesting between 1/n^1 and 1/n^2?
    Sure.
     
  7. May 19, 2013 #6
    Alright so I say that since 1/n in DIvergent and 1/n^2 is CONvergent, the sigma sigma overall is DIvergent?

    Not sure, 1/n^1.5?
     
  8. May 19, 2013 #7

    mfb

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    Right.
    That is a good idea.
     
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