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Infinite series Σ 1/(ln(e^n+e^-n)) = Σ 1/ln(n)?

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove whether [tex]\sum \frac{1}{ln(e^{n}+e^{-n})}[/tex] converges or diverges


    2. Relevant equations



    3. The attempt at a solution

    (second post today... sorry, I just want to make sure I'm getting this right)


    Since [tex]e^{n}+e^{-n}[/tex] goes to infinity as n goes to infinity, could I say that [tex]\sum \frac{1}{ln(e^{n}+e^{-n})}[/tex] is like [tex]\sum \frac{1}{ln(n)}[/tex]?

    I know that [tex]\sum \frac{1}{ln(n)}[/tex] is definitely divergent because it is > 1/n and I can use the comparison test.

    But could I argue that [tex]\sum \frac{1}{ln(e^{n}+e^{-n})}[/tex] and [tex]\sum \frac{1}{ln(n)}[/tex] are essentially the same?
     
  2. jcsd
  3. Oct 22, 2009 #2
    No, you can't say that... for example n^2 goes to infinity as n goes to infinity, sum of 1/n diverges and sum of 1/n^2 converges.

    But you can notice that [tex]e^n + e^{-n} < e^n + e^n[/tex] ...
     
  4. Oct 22, 2009 #3
    Right, so [tex]\frac{1}{ln(e^{n}+e^{-n})} > \frac{1}{ln(e^{n}+e^{n})} = \frac{1}{ln(2)+ln(e^{n})} = \frac{1}{ln(2)+n}[/tex]

    [tex]\sum \frac{1}{ln(2)+n}[/tex] certainly looks like it diverges... but how would I prove that?

    comparison test wouldn't work since it's not quite true that [tex]\sum \frac{1}{ln(2)+n} > 1/n[/tex]
     
  5. Oct 22, 2009 #4
    If you could show that ∑ 1/ln(en + e-n) is greater than ∑ 1/ln(n) then you could compare them and show that the first one diverges. Otherwise, it won't help you.

    Try another comparison with en + e-n < e2n
     
  6. Oct 22, 2009 #5
    Does [tex]\sum \frac{1}{n+1} = 1/2 + 1/3 + 1/4 + ... [/tex] converge or diverge?
     
  7. Oct 22, 2009 #6
    n + 1 < 2n
    1/(n + 1) > 1/(2n)

    Σ 1/(2n) = (1/2)Σ 1/n diverges, so Σ 1/(n + 1) diverges.
     
  8. Oct 22, 2009 #7
    Exactly, and also Σ 1/(n+1) = (Σ 1/n) - 1, another way to see that it diverges. And 1/(ln(2)+n) > 1/n+1.
     
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