Question

Use a comparison to determine whether the integral converges or diverges.

Answer 1

Comparison Theorem?

Answer 2

ooh okay ?

Answer 3

Oh I was asking which test you'd prefer using :)

Answer 4

ohh lol either one

Answer 5

so?

Answer 6

Attempting to solve it :)

Answer 7

ok

Answer 8

Bah, I need surji here. @surjithayer

Answer 9

Oh wait, I think i've got it. Have you tried solving it as a series yet?

Answer 10

nope

Answer 11

Would you happen to have the answer for this question? :\

Answer 12

@Loser66 and @terenzreignz can u help? :/

Answer 13

I didn't know you could use comparison with integrals... :D

Answer 14

improper integral? use lim?

Answer 15

Yeah, you can rewrite it as a series

Answer 16

I guess the same principles as in series applies...
\[\Large \frac{x}{1+x^3} < \frac{x}{x^3}=\frac1{x^2}\]

I guess that's the framework of it, @Jhannybean ?

Answer 17

Yep.what I was going to explain haha you're right.

Answer 18

And \[\frac1{x^2}\] is convergent, since we can sorta reverse the integral test (in that if the integral converges, so does the series)

In this case, the series
\[\Large \frac1{n^2}\] is convergent, then so is the improper integral (to infinity) of \[\Large \frac1{n^2}\]

Answer 19

crud, I meant the integral of
\[\Large \frac1{x^2}\]

Answer 20

and because \[\frac{ 1 }{ x^2 } \] converges by the p-series test, by comparing an and bn, you can see that an < bn, \[\sum_{n=1}^{\infty}\frac{ 1 }{ n^2 }\] is convergent

Answer 21

Not forgetting of course, that
\[\Large 0 < \frac{x}{1+x^3} \]for as long as\[\Large x \in [1,\infty)\]

Answer 22

wow okay

Answer 23

Therefore by the comparison test, \[\int\limits_{1}^{\infty}\frac{ x }{ 1+x^3 }=\sum_{n=1}^{\infty}\frac{ x }{ 1+x^3 }\] and \[a _{n}=\frac{ x }{ 1+x^3 } , b _{n}=\frac{ 1 }{ n^3 }\]

Answer 24

we can compare the two and say that because \[\sum_{n=1}^{\infty}\frac{ 1 }{ n^3 }\] converges by p-series, \[\sum_{n=1}^{\infty}\frac{ x }{ 1+x^3 }\] will converge also

Answer 25

Thanks for the jump start, @terenzreignz :) I just had this on my final and momentarily forgot!!

Answer 26

Just a reminder...
\[\int\limits_{1}^{\infty}\frac{ x }{ 1+x^3 }=\sum_{n=1}^{\infty}\frac{ x }{ 1+x^3 }\]
Not true~

The integral does not necessarily equal the infinite series, although their convergences are tied... IE, they have to either both converge or both diverge.

Answer 27

By the limit comparison, no? Oh..

Answer 28

Limit comparison only occurs between two series (or two integrals, that I don't know, but I guess it can be shown to hold)

But yeah, improper integral is not equal to the infinite series...

Improper integral would be the area under that curve, but...