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Does this series converge

  1. Apr 6, 2009 #1
    I have a problem.
    Does the series 1/(3n-2)(3n+1) converge or diverge. If it converges find its sum.
    How would I do this.
  2. jcsd
  3. Apr 6, 2009 #2
    Also, does the series ln(1-1/n^2) converge or diverge. If so calculate the sum.
    I know to first calculate the partial sum but i don't know how to do this. Someone please help.
  4. Apr 6, 2009 #3
    The first series converges for sure. To see this, simply note that it is asymptotically less than 1/n^2, a convergent series.

    If the second one starts at n=1, then I'm sure it doesn't converge. Plugging in n=1, you get that the first term is negative infinity.

    To evaluate the first series... well, that's a good question.
  5. Apr 7, 2009 #4
    Observe that

    [tex]\frac{1}{(3n-2)(3n+1)} = \frac13 \left( \frac{1}{3n-2} - \frac{1}{3n+1} \right)[/tex]

    so you get a telescoping series. Similarly,

    [tex]\ln \left( 1 - \frac{1}{n^2} \right) = \ln \frac{(n - 1)(n + 1)}{n^2}
    = \ln (n - 1) - 2 \ln n + \ln (n + 1)[/tex]

    which also collapses on summation. (Of course, you must start at n = 2.)
    Last edited: Apr 7, 2009
  6. Apr 10, 2009 #5
    One more problem.
    How do I calculate the Nth partial sum to the series 1/(sqrt(n)+sqrt(n+1)) and hence determine if this series converges.
    And determine if the series sin^2(1/n) diverges or converges.
    Someones help would be appreciated.
    I'm not to sure how to start either of them.
  7. Apr 10, 2009 #6
    For the first one, you should find that the first few partial sums

    [tex]\sum_{n=0}^{N} \frac{1}{\sqrt{n} + \sqrt{n+1}}[/tex]

    are [tex]1, \sqrt{2}, \sqrt{3}, 2, \sqrt{5}[/tex]. Can you make a conjecture from this and prove it (by induction)?

    For the second one, compare with 1/n2.
    Last edited: Apr 10, 2009
  8. Apr 12, 2009 #7
    Why do I prove it by induction?
    And I'm still not sure how to find the Nth sum. What is the meaning of the Nth sum?
  9. Apr 12, 2009 #8
    If you have a series
    [tex]\sum_{n=0}^{\infty} a_n,[/tex]
    the Nth partial sum is
    [tex]\sum_{n=0}^N a_n.[/tex]

    As for proving that, I'm just suggesting a way to prove it, that's probably the easiest.
  10. Apr 12, 2009 #9
    Okay, I see now. Thanx.
    So how would I apply it to the example 1/(sqrt(n)+sqrt(n+1))?
  11. Apr 12, 2009 #10
    Well, as I suggested, your conjecture should be that

    [tex]\sum_{n=0}^{N} \frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{N + 1}.[/tex]

    You would prove that by induction on N. (Have you done induction proofs before? Any good analysis course should teach it.) Clearly it's true if N = 0; now suppose it's true for N - 1, that is, suppose that

    [tex]\sum_{n=0}^{N-1} \frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{(N - 1) + 1} = \sqrt{N}.[/tex]


    \sum_{n=0}^{N} \frac{1}{\sqrt{n} + \sqrt{n+1}}
    &= \sum_{n=0}^{N-1} \frac{1}{\sqrt{n} + \sqrt{n+1}} + \frac{1}{\sqrt{N} + \sqrt{N+1}} \\
    &= \sqrt{N} + \frac{1}{\sqrt{N} + \sqrt{N+1}},

    and all you have to do now is prove that this equals [tex]\sqrt{N + 1}[/tex].
  12. Apr 12, 2009 #11
    It goes from n=1 to infinity.
  13. Apr 12, 2009 #12
    Then just subtract the n = 0 term from everything, so

    [tex]\sum_{n=1}^{N} \frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{N + 1} - 1.[/tex]
  14. Apr 12, 2009 #13
    Thanks heaps for that.
    How would I prove that this converges then.
    Do I do a comparison test with sqrt(N+1)-1 and 1/(sqrt(n)+sqrt(n+1))
  15. Apr 14, 2009 #14
    To see if the series sin^2(1/n) converges or diverges do I do a comparison test with 1/n^2.
  16. Apr 15, 2009 #15
    No; you'd use the definition of convergence of a series.
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