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Sum convergence - is my approach flawless?

  1. Jan 18, 2005 #1
    I have to find out, whether this sum converges:

    \sum_{n = 2}^{+ \infty} n^{\alpha} \left( \log (n+1) - \log n \right)^4

    So I rewrote it:

    \sum_{n = 2}^{+ \infty} n^{\alpha} \left( \log \left(1 + \frac{1}{n} \right) \right)^4 =

    \sum_{n = 2}^{+ \infty} n^{\alpha - 4} \left[ \frac {\log \left(1 + \frac{1}{n} \right)}{\frac{1}{n}} \right]^4

    Now I can see that the expression with log goes to 1, so the sum gets reduced to

    \sum_{n = 2}^{+ \infty} \frac {1}{n^{4 - \alpha}}

    and the result is obviously [itex]\alpha \in \left(0, 3\right)[/itex].

    But, I don't know how exactly should I justify my decision. I'm not talking about the very last step now, but the one in which I threw away the logarithm because of the limit 1. We have to give reasons for each non-trivial step we do, so I think I should write some theorem or rule under that step...

    Thank you.
  2. jcsd
  3. Jan 18, 2005 #2


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    If you're saying that:

    [tex]\sum_{n = 2}^{+ \infty} n^{\alpha - 4} \left[ \frac {\log \left(1 + \frac{1}{n} \right)}{\frac{1}{n}} \right]^4=\sum_{n = 2}^{+ \infty} \frac{1}{n^{4-\alpha}}[/tex]
    ..then no.
    But you can use it to compare the original series it with the right hand side.

    If you can show
    [tex]\left[ \frac {\log \left(1 + \frac{1}{n} \right)}{\frac{1}{n}} \right]^4<1[/tex]
    and approaches 1, then you're home free. That means the original series converges too.
  4. Jan 18, 2005 #3
    Yes, I see what you mean. But...how should I prove this:

    If you can show
    [tex]\left[ \frac {\log \left(1 + \frac{1}{n} \right)}{\frac{1}{n}} \right]^4<1[/tex]

    I only know the limit is 1, nothing more...
  5. Jan 18, 2005 #4


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    Use the limit comparison test. You know that [tex]n^{\alpha-4} [/tex] converges for [tex]\alpha <= 2 [/tex].

    Compare your given series

    [tex]\sum_{n = 2}^{+ \infty} n^{\alpha - 4} \left[ \frac {\log \left(1 + \frac{1}{n} \right)}{\frac{1}{n}} \right]^4 [/tex]

    to the series [tex]n^{\alpha-4} [/tex], and calculate the limit... which you've already discovered is 1 (this is essentially what you did anway). Since this is finite and greater than zero, your series must converge.

    This also shows that when [tex]n^{\alpha-4} [/tex] diverges ( [tex]\alpha > 2[/tex], your series also diverges.
    Last edited: Jan 18, 2005
  6. Jan 18, 2005 #5
    Now it's clear, thank you learningphysics and Galileo. Limit comparsion test is the clue.
  7. Jan 22, 2005 #6
    I thought it was clear to me, but some time ago I had to solve similar problem and I realized that comparing my original sum to some other sum only tells me, for which [itex]\alpha[/itex] it converges surely, but it doesn't tell me all [itex]\alpha[/itex] values for which it converges. I hope I expressed it in an understandable way...
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