Is $$\sum_{n=3}^\infty \frac{1}{n^2 \log^3 n}$$ absolutely convergent?

Using comparison test, since $n$ is greater than or equal to 3: $$\frac{1}{n^2 \log^3 n} \leq \frac{1}{n^2}$$

And, we know $$\sum_{n=3}^\infty \frac{1}{n^2}$$ converges by p-test.

Therefore, $$\sum_{n=3}^\infty \frac{1}{n^2 \log^3 n}$$ converges absolutely.


You've answered your own question. Your proof is correct.

Note that in this case "absolutely convergent" is just the same as "convergent" since all the terms are positive.

  • $\begingroup$ oops i was suppose to ask if what I did is right or not. what about for conditionally convergent? $\endgroup$ – Risa Jun 11 '13 at 4:40
  • $\begingroup$ A conditionally convergent series is simply a convergent series which does not converge absolutely. Since this series does converge absolutely, it cannot converge conditionally. The two are mutually exclusive. (In general, any series is exactly one of (1) divergent, (2) conditionally convergent, and (3) absolutely convergent.) $\endgroup$ – 6005 Jun 11 '13 at 4:45

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