Question

Does this series converge or diverge?

Answer 1

Well, limit comparison seems to work. You know limit comparison test?

Answer 2

yes
but what can I compare it to?

Answer 3

1/n

Answer 4

Apparently, I can't compare it to 1/j or 1/(j^1/3)

Answer 5

Why cant you?

Answer 6

The answer says I have to compare it to 1/ (2j)

Answer 7

You dont HAVE to compare it to something, its just a suggestion.

Answer 8

Theres no real strict requirements for what you choose as a comparison, as long as the comparison helps you out and can be used in your problem.

Answer 9

but when i use 1/j as my comparison, I get converges, when the answer should be diverges.

Answer 10

You shouldnt get converge. I mean the limit comparison. For the limit comparison, you take your original series and divide it by a comparison of your choice. If the division of the original and your comparison has a positive and finite limit as n goes toinfinity, then both series either converge OR diverge. The reason why 1/j is so convenient is because its a series we already know DIVERGES. So when I choose it as a comparison, I know its behavior, all I need is for the limit comparison to give us a finite and positive answer. So using 1/j, I get:

\[\frac{ 1 }{ j + j^{1/3} }\div \frac{ 1 }{ j } = \frac{ j }{ j + j^{1/3} }\]
\[\lim_{j \rightarrow \infty}\frac{ j }{ j+ j^{1/3} }= 1\]Therefore either both series converge or both diverge via the limit comparison test. But since we already know 1/j diverges, our problem also must diverge.

Answer 11

I'm not sure if I'm supposed to use that, since I don't think I learned it...The answer says:

Answer 12

Comparison is vague as there are direct comparison and limit comparison tests. My old textbook introduces them in the same chapter.

Answer 13

so what is this answer telling me?

Answer 14

Well, the comparison portion is vague, it could mean one of two. We could always try the integral test and do it a way we know you are supposed to know if youd like.

Answer 15

ok

Answer 16

So for the integral test, the function must be positive and continuous, which is np. Then the function must be decreasing for some j greater than 0. So that we means we take the derivative and see that its decreasing. Since your answer says integral test works, we can assume the derivative is definitely negative and decreases. So now we need to do this integration:

\[\int\limits_{1}^{\infty}\frac{ 1 }{ j + j^{1/3} }dj\]Thinkyou could integrate that improper integral?

Answer 17

no...ive tried...

Answer 18

Alright, ill give it a shot, Ihave an idea on how to do it.

Answer 19

I apologize. Got busy and not the easiest integral, lol.
u = j^(1/3)
j = u^3
du = 1dj/3j^(2/3)
dj = 3j^(2/3)*du
\[\int\limits_{}^{}\frac{ 3u^{2} }{ u^{3}+u }du \implies 3\int\limits_{}^{}\frac{ u }{ u^{2}+1 }du\]
v = u^2 + 1
dv = 2udu
du = dv/2u
\[3\int\limits_{}^{}\frac{ u }{ v }*\frac{ dv }{ 2u } \implies \frac{ 3 }{ 2 }\int\limits_{}^{}\frac{ 1 }{ v }dv= \frac{ 3 }{ 2 }\ln|v| \implies \frac{ 3 }{ 2 }\ln|j^{2/3}+1| \] and that integral with limits from 1 to infinity diverges.