Question

This is a calc 2 question, absolute convergence.
Im not sure what tests for convergence to use on
Series as n=1 to infinity of 2^n(lnn)*1/n.
Is this series, divergent, absolutely or conditionally convergent, explain your reasoning.

Answer 1

\[\sum_{n=1}^{\infty}\frac{ 2^n(\ln(n)) }{ n }\]

Answer 2

my firs thought was to try to put it in the form of a geometric series, but im not sure if that's right because of the n on the bottom

Answer 3

Then I thought maybe integral test, but the "n" power, baffles me

Answer 4

unfortunately LaTeX doesn't work on my windows partition, I will have a look at this in a few though if it remains unanswered.

Answer 5

That's okay @Spacelimbus, could you see whenever, if I get it, if I use the limit comparison test, can I compare it to the series 1/n. taking the lim of a_n/b_n, the n's cancel out, and I would have a limit =infinity.

Answer 6

and therefore divergent

Answer 7

I believe if you want to do any kind of limit comparision test, then you most likely want to drop the ln(n) term, because it doesn't take on too large values

Answer 8

but your idea seems to work out too, your reasoning.

Answer 9

yes, just to say the same thiing as what you are saying, that's why I compared, the a_n to a b_n = 1/n , then did limit comparison. yayy I get it. thx

Answer 10

well done