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Determine if the series $\frac{((\ln(n))^3}{n}$ is convergent or divergent.

What I Tried :- If I were to use the comparison test would I end up with $(\ln(n))^3 > 1/n^2 > 0$. So $\frac{1}{n^2}$ is convergent by $p$-test as $(p=2>1)$. Therefore the original series is convergent by comparison test.

Can anyone help me understand If I am heading in the right direction?

Anonymous
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MomLearningMath
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1 Answers1

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That does not work.

The comparison test states that if $\sum b_{k}$ converges and $0\leq a_{n} \leq b_{n}$ for all natural numbers $n$ greater than some fixed natural number $N$, then $\sum a_{k}$ converges.

On the other hand if $\sum a_{k}$ diverges and $0\leq a_{m} \leq b_{m}$ for all natural numbers $m$ greater than some fixed natural number $M$, then $\sum b_{k}$ diverges.

Hint: What happens if you compare $\frac{(\ln n)^{3}}{n}$ with $\frac{1}{n}$

  • How do we know the 1/n is what we want to compare the original to? Figuring out b_n is the most confusing part for me on direct comparisons. – MomLearningMath Oct 03 '20 at 06:52
  • $\frac{(\ln n)^{3}}{n}$ is essentially $\frac{\text{Something that gets larger}}{n}$, and we know that $\sum \frac{1}{n}$ diverges. So it is natural to ask if "Something that gets larger" is greater than $1$ from some point onwards –  Oct 03 '20 at 16:19