# Is the series conditionally convergent, absolutely convergent or divergent $\sum(-1)^n\frac{\ln^3 n}n$.

Determine if the series

$$\sum_{n=1}^\infty(-1)^n\frac{\ln^3 n}n$$

is conditionally convergent, absolutely convergent or divergent.

By comparison test I got $$a_n > b_n$$ therefore divergent with $$a_n=\frac{\ln^3 n}n$$ and $$b_n=\frac 1n$$.

Then with the alternating test I concluded it converges.

Therefore the original series is conditionally convergent.

Is my though process correct here or is there correction needed?

• What are $a_n$ and $b_n$? – José Carlos Santos Oct 3 '20 at 14:08
• a_n: (ln (n))^3 / n. b_n: 1/n – MomLearningMath Oct 3 '20 at 14:13
• Yes. You are correct. – Mark Viola Oct 3 '20 at 14:47

Yes your though process is correct since eventually $$\log^3 n \ge 1$$ for the absolute series we have
$$\frac{\ln^3 n}n \ge \frac1n$$
$$f(x)=\frac{\ln^3 x}x \implies f'(x)=-\frac{(\log x -3)\log^2 x}{x^2}<0$$