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Absolute Convergent, Conditionally Convergent?

  1. May 13, 2012 #1
    1. The problem statement, all variables and given/known data
    ##\sum _{n=2}\dfrac {\left( -1\right) ^{n}} {\left( \ln n\right) ^{n}}##


    3. The attempt at a solution
    I have applied the Alternating Series test and it shows that it is convergent. However, I need to show that it's either absolute conv. or conditionally conv.

    Next, I tried the root test:
    ##\lim _{n\rightarrow \infty }\sqrt {\left| \left( \dfrac {1} {\ln n}\right) ^{n}\right| }##, **Correction; this is the root of n**
    Now, I'm tempted to use direct comparison with 1/n harmonic series and show divergence. However, I don't know, if I am allowed to do this since the root test tells me that then I have to find the limit of my An and classify it accordingly.
    According to the root test:
    if the limit >1 or infinity it diverges
    if the limit <1 it converges absolutely.
    if the limit =1 it's inconclusive.
     
  2. jcsd
  3. May 13, 2012 #2

    LCKurtz

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    What do you get when you simplify$$
    \lim_{n\rightarrow\infty}\root\scriptstyle ^n\of {\frac 1 {(\ln n)^n}}$$Then what happens as ##n\rightarrow \infty##? Don't answer that by talking about some series test. This is just the limit of a sequence. What do you get?
     
  4. May 13, 2012 #3
    $$
    \lim_{n\rightarrow\infty}\root\scriptstyle ^n\of {\frac 1 {(\ln n)^n}}$$ =0.

    Then, by the root test my series converges absolutely.
     
  5. May 13, 2012 #4

    LCKurtz

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    Do you have an aversion to showing your work? How did you get zero?
     
  6. May 13, 2012 #5
    ##\lim _{n\rightarrow \infty }\left( \dfrac {1} {\ln n}\right) ^{\dfrac {n} {n}}=\lim _{n\rightarrow \infty }\dfrac {1} {\ln n}=0##
     
  7. May 13, 2012 #6

    LCKurtz

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    Yes, thank you. So you see the root test was perfect for this problem. Quick and easy and instantly gives convergence. Nothing else needed.
     
  8. May 13, 2012 #7
    Thank you.
     
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