# Determining convergence of series

#### calcboi

##### New member
I have a question on which test to use for series n=1 to infinity for (-1)^n / (n^3)-ln(n) in order to determine convergence/divergence. I am pretty sure I determined it converges through the Alternating Series Test(correct me if I'm wrong) but I am not sure whether it is conditional or absolute. I tried the Direct Comparison Test but it was inconclusive, and I am stuck now on what to do. I also tried Limit Comparison but the limit goes to infinity so it is also inconclusive. Can you please help?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Yes you are correct.

near infinity the term n^3 is dominant over ln (n).

#### calcboi

##### New member
BTW, I used the comparison c(x) = 1/n^3 for DCT and LCT since we know 1/n^3 converges by p-series.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
BTW, I used the comparison c(x) = 1/n^3 for DCT and LCT since we know 1/n^3 converges by p-series.
Right, $\displaystyle\lim_{n\to \infty}\left(\frac{1}{n^3-\log n}:\frac{1}{n^3}\right)=\ldots =1\neq 0$, so the series is absolutely convergent.

#### calcboi

##### New member
What test did you use to determine absolute convergence? Or was that just analyzing end behavior?

MHB Math Helper

#### Fernando Revilla

##### Well-known member
MHB Math Helper
When I tried the Limit Comparison Test, I got infinity as n approaches infinity. How did you get 1?
$\displaystyle\lim_{n\to \infty}\left(\frac{1}{n^3-\log n}:\frac{1}{n^3}\right)=\lim_{n\to \infty}\frac{n^3}{n^3-\log n}=\lim_{n\to \infty}\frac{1}{1-(\log n/n^3)}=\frac{1}{1-0}=1$