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Series convergence / divergence

  1. Jun 7, 2014 #1
    1. The problem statement, all variables and given/known data

    Does the following series converge or diverge? If it converges, does it converge absolutely or conditionally?
    [itex]\sum^{\infty}_{1}(-1)^{n+1}*(1-n^{1/n})[/itex]

    2. Relevant equations
    Alternating series test

    3. The attempt at a solution
    I started out by taking the limit of ##a_n: Lim_{n\rightarrow\infty}(1-n^{1/n})##=1-1=0.
    So via the alternating series test, the original series converges.

    Next, I have to figure out whether it converges absolutely or conditionally, and this is where I'm stuck. I suppose I have to first find out whether the term ##a_n##, which is ##(1-n^{1/n})##, diverges or converges. But what test do I use for this? The limit test is silent because as I found above, the limit is zero. I've tried ratio test and limit comparison test to no avail. Root test doesn't seem to lead anywhere. It's not a geometric or telescoping series so those options are out. Maybe I could use comparison test, but I don't know what term I would use for comparison. Integral test seems really difficult. So I'm stuck and would appreciate any helpful pointers.
     
  2. jcsd
  3. Jun 7, 2014 #2

    lurflurf

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    think about [(n^(1/n)-1)/(1/n)]/n
    compare to log(n)/n
     
  4. Jun 7, 2014 #3
    Wait, so...I gather you're saying that ##\sum(-1)^{n+1}*(1-n^{1/n})=\sum(-1)^n*(n^{1/n}-1)##, so when we consider the positive term ##(n^{1/n}-1)##, it can be re-written as:

    ##\frac{(n^{1/n}-1)*n}{n}##, aka:


    ##\frac{\frac{(n^{1/n}-1)}{\frac{1}{n}}}{n}##?

    And from here we do a comparison test with ##\frac{\ln n}{n}##? OK, From punching some numbers into the calculator, I can see that ##\frac{\frac{(n^{1/n}-1)}{\frac{1}{n}}}{n} > \frac{\ln n}{n}##, so via the comparison test the positive term diverges, meaning the original alternating series converges conditionally?

    But if so, then I'm not sure how to show rigorously that ##\frac{\frac{(n^{1/n}-1)}{\frac{1}{n}}}{n} > \frac{\ln n}{n}##, other than to say it seems to be true for some arbitrarily large n...am I on the right track here?
     
  5. Jun 7, 2014 #4

    lurflurf

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    ^Exactly
    n^(1/n)-1>=0 for all n (n=1,2,3,4,...)
    x>=log(1+x) for all x (x>=0)
    thus
    n^(1/n)-1>log{1+[n^(1/n)-1]}=log(n)/n
    ∑log(n)/n diverges so too does ∑[n^(1/n)-1]
     
  6. Jun 7, 2014 #5
    You can use derivatives to show this (increasing/decreasing functions).
     
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