Welcome to our community

Be a part of something great, join today!

Determining convergence of series

calcboi

New member
Mar 31, 2013
16
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
Jan 17, 2013
1,667
Yes you are correct.

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

calcboi

New member
Mar 31, 2013
16
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
Jan 29, 2012
661
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
Mar 31, 2013
16
What test did you use to determine absolute convergence? Or was that just analyzing end behavior?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

calcboi

New member
Mar 31, 2013
16

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
When I tried the Limit Comparison Test, I got infinity as n approaches infinity. How did you get 1?
$\displaystyle\lim_{n\to \infty}\left(\frac{1}{n^3-\log n}:\frac{1}{n^3}\right)=\lim_{n\to \infty}\frac{n^3}{n^3-\log n}=\lim_{n\to \infty}\frac{1}{1-(\log n/n^3)}=\frac{1}{1-0}=1$