Question

@Kainui

Answer 1

this question asks to "use any method to determine whether the series converges"

Answer 2

ooops i entered the wrong problem! let me fix that

Answer 3

Tried ratio test ?

Answer 4

Alright so we can use any method, which ones have you tried and/or are considering?

Answer 5

@hartnn yes! i was actually wondering if that was the best choice. however i am trying to work it out now and currently have...
\[\lim_{k \rightarrow \infty}\frac{ \ln k+1 }{ e \ln k }\]

Answer 6

my book says converges but I'm not sure if I'm on the right track ..?

Answer 7

When they say any method I just instantly thought of doing this... hehe... \[\sum_{k=1}^{\infty}\ln(k^{e^{-k}})=\ln (\prod_{k=1}^{\infty}k^{e^{-k}})\] and just playing around with that. But sorry I'm only confusing things playing around.

Answer 8

@Kainui im afraid i haven't learned that yet lol

Answer 9

That's ok, it's not important just turned it into an infinite product since log rules do that...

So you are doing good with the ratio test, continue with it! Right now you have infinity/infinity if you take the limit which is an indeterminate form right?

Answer 10

hmm so from this point couldn't i just take the derivative of the numerator and denominator??

Answer 11

Exactly!

Answer 12

if you can use L'Hopital's rule :)

Answer 13

yes @hartnn ! however if i take the derivative of the denominator wouldn't that be [zero * (1/k)]

Answer 14

e is constant, it can be taken out of derivative
\((e\ln k)' = e (\ln k)'\)

Answer 15

oh!!! thats right !

Answer 16

=)

Answer 17

didn't catch that :)

Answer 18

see if you get the limit as 1/e :)

Answer 19

indeed i did :)

Answer 20

Eureka! :)

Answer 21

need i say that since 1/e <1 , by ratio test, that converges ?
or you figured that out on your own ?

Answer 22

got it :) Thank you both!