Convergence of ln(k)/k^3 Series

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Homework Statement



Determine whether the series converges or diverges.

\sum \frac{lnk}{k^3}

Homework Equations


The Attempt at a Solution



Since lnk always less than 0, so \frac{lnk}{k^3} \leq \frac{1}{k^3} and \frac{1}{k^3}diverges so\frac{lnk}{k^3} diverges.
 
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zeion said:

Homework Statement



Determine whether the series converges or diverges.

\sum \frac{lnk}{k^3}


Homework Equations





The Attempt at a Solution



Since lnk always less than 0, so \frac{lnk}{k^3} \leq \frac{1}{k^3} and \frac{1}{k^3}diverges so\frac{lnk}{k^3} 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.
 
Can I argue that ln(k) &lt; k then \frac{ln(k)}{k^3} &lt; \frac{k}{k^3}then it converge because \frac{1}{k^2} converges?
 
I can buy that. However, your should be able to convince yourself that ln(x) < x.
 
Yeah since ln(x) is the inverse of e^x and they reflect on the y =x.
 
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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