- #1
IntegrateMe
- 217
- 1
Determine if the series converges absolutely, conditionally, or if it diverges. Give justifications for your answers.
(note: "E" is the sigma notation)
1. E from 1 to infinity of ln(n)/n3
Ok, i basically used the limit comparison test and pulled out a 1/n3. As we know by p-series, 1/n3 converges and now becomes our "bn."
So, using the LCT i get... limit going to infinity of an/bn = lim to infinity of [ln(n)/n3]/(1/n3) which results in the lim to infinity of ln(n) which = infinity. Because our limit is infinity it means that if bn diverges, an diverges as well. But in this case, bn converged! Any thoughts??
Because bn converged i thought the answer would be "converges conditionally" but i was marked points off because the answer is "converges absolutely."
(note: "E" is the sigma notation)
1. E from 1 to infinity of ln(n)/n3
Ok, i basically used the limit comparison test and pulled out a 1/n3. As we know by p-series, 1/n3 converges and now becomes our "bn."
So, using the LCT i get... limit going to infinity of an/bn = lim to infinity of [ln(n)/n3]/(1/n3) which results in the lim to infinity of ln(n) which = infinity. Because our limit is infinity it means that if bn diverges, an diverges as well. But in this case, bn converged! Any thoughts??
Because bn converged i thought the answer would be "converges conditionally" but i was marked points off because the answer is "converges absolutely."