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Math Help - Absolute or conditional convergence

  1. #1
    Senior Member mollymcf2009's Avatar
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    Absolute or conditional convergence

    Ok, so I know that this series converges, but can someone please show me how I would determine absolute or conditional? Thanks!

    \sum^{\infty}_{n=3}  (-1)^{n-1} \frac{ln(n)}{n}
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    Does the sequence \frac{\ln (n)}{n} decrease and converge to zero?
    That is the alternating series test.

    Now does the series \sum\limits_{n = 3}^\infty  {\frac{{\ln (n)}}{n}} converge?
    If not then the convergence is not absolute.
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by Plato View Post
    Does the sequence \frac{\ln (n)}{n} decrease and converge to zero?
    That is the alternating series test.

    Now does the series \sum\limits_{n = 3}^\infty  {\frac{{\ln (n)}}{n}} converge?
    If not then the convergence is not absolute.
    I understand all that. What I am trying to figure out is since I have shown that \sum\limits_{n = 3}^\infty  {\frac{{\ln (n)}}{n}} does indeed converge. Does it converge conditionally because of the (-1)^{n-1}? This was a question on a test that I had this morning. It was a two part question. The first part I had to show what test I would use to figure out if it converged or not. I used the Alternating Series test and found that b_n = 0, a_{n+1} < a_n, and that it was alternating, which proves that the series converges. For part b, it asked me to show if it converged conditionally or absolutely. I said it converged conditionally. It converged because it met the criteria for an alternating series. However, with (-1)^{n-1} I said that by having this as part of the series, it would affect its convergence because it caused the values to bounce around and not hone in on a specific value. Is that right or no?

    Thanks for the response!
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    Quote Originally Posted by mollymcf2009 View Post
    I understand all that. What I am trying to figure out is since
    I have shown that \color{red}\sum\limits_{n = 3}^\infty  {\frac{{\ln (n)}}{n}} does indeed converge.
    How could you understand all this if you think you have proved the above in red.
    The fact is: \sum\limits_{k = 3}^\infty  {\frac{{\ln (k)}}{k}} diverges.

    It should have been worked out in the textbook.

    Just note that k \geqslant 3\, \Rightarrow \,\frac{{\ln (k)}}{k} > \frac{{k - 1}}{{k^2 }}.
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  5. #5
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by Plato View Post
    How could you understand all this if you think you have proved the above in red.
    The fact is: \sum\limits_{k = 3}^\infty  {\frac{{\ln (k)}}{k}} diverges.

    It should have been worked out in the textbook.

    Just note that k \geqslant 3\, \Rightarrow \,\frac{{\ln (k)}}{k} > \frac{{k - 1}}{{k^2 }}.
    Wow, ok, sorry. I guess I may have left something out of the question, it is just from memory. I do understand this concept, I just wanted some clarification on the process of conditionally vs. absolute. I'll see if I can clarify the actual problem with my professor and re post.
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  6. #6
    Senior Member mollymcf2009's Avatar
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    Clarified question/answer

    Ok, so I got a response back from my professor. It is true that the series \frac{ln(n)}{n} diverges. But with the (-1)^{n-1} it converges conditionally. I was confused, because I just thought that if a series is divergent, it's divergent. Period. I think I just assumed that for a series to be conditionally convergent, it began as convergent and due to making a change to it i.e. (-1)^{n-1} made it be conditionally convergent. Anyway, looks like I solved it correctly on my test, but didn't look at the series long enough to realize that it was actually divergent.
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