Series converge or diverge

  • Thread starter zeion
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  • #1
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Homework Statement



Determine whether the series converges or diverges.

[tex] \sum \frac{lnk}{k^3} [/tex]


Homework Equations





The Attempt at a Solution



Since lnk always less than 0, so [tex] \frac{lnk}{k^3} \leq \frac{1}{k^3}[/tex] and [tex]\frac{1}{k^3} [/tex]diverges so[tex] \frac{lnk}{k^3}[/tex] diverges.
 

Answers and Replies

  • #2
34,896
6,638

Homework Statement



Determine whether the series converges or diverges.

[tex] \sum \frac{lnk}{k^3} [/tex]


Homework Equations





The Attempt at a Solution



Since lnk always less than 0, so [tex] \frac{lnk}{k^3} \leq \frac{1}{k^3}[/tex] and [tex]\frac{1}{k^3} [/tex]diverges so[tex] \frac{lnk}{k^3}[/tex] diverges.
I count three mistakes here.
1. Why do you think that ln(k) is always < 0? I'm assuming that k = 1, 2, 3, ...
2. The series whose general term is 1/k3 converges.
3. If you want to show that a given series diverges, its terms must be larger than those of a divergent series. If the terms of the given series are smaller than those of a divergent series, you can't conclude anything.
 
  • #3
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Can I argue that [tex]ln(k) < k [/tex] then [tex] \frac{ln(k)}{k^3} < \frac{k}{k^3} [/tex]then it converge because [tex]\frac{1}{k^2} [/tex] converges?
 
  • #4
34,896
6,638
I can buy that. However, your should be able to convince yourself that ln(x) < x.
 
  • #5
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Yeah since ln(x) is the inverse of e^x and they reflect on the y =x.
 
  • #6
34,896
6,638
The fact that y = ln(x) and y = ex are reflections across the line y = x doesn't prove that ln(x) < x. If y = ex happens to cross this line a few times means that ln(x) will do so, also.
 

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