Question

determine if the improper integral converges...

Answer 1

if it does, evaluate it \[\int\limits_{2}^{\infty} \frac{ 1 }{ \ln(x) } dx\]

Answer 2

if the series converges, so does the integral
and vice versa

Answer 3

Would \(\displaystyle \sum_{n=2}^{\infty}\frac{1}{\ln x}\) converge?

Answer 4

Well, you know harmonic series would diverge aint it so?

and,

\(\displaystyle \sum_{n=2}^{\infty}\frac{1}{\ln x}>\sum_{n=2}^{\infty}\frac{1}{x}\)

Answer 5

i thought it would comverge because lim as x approaches infinity = 1/infinity = 0

Answer 6

that is just for divergence

Answer 7

if lim = 0, you can't make conclusions,
if lim =/ 0, then for sure diverges

Answer 8

for x > 2
1/(xlnx) < 1/ ln x
therefore
∫ 1/(xlnx) < ∫ 1/ ln x

the left side can be integrated, it goes to infinity. so the right side must go to infinity

Answer 9

yeah, or that, comparing to x ln(x) or just x ... doesn't matter:)

Answer 10

((you could ideally compare to any x^p: p>0 from some n=j an on (forever), and convergence from n=j would dictate convergence from n=2, since there aren't any singular points over [2,∞) ....)

anyway, I got to get back to my stuff.
Sorry if I interrupted.

Answer 11

im kinda confused...it diverges?

Answer 12

Yes

Answer 13

let's clarify

Answer 14

\(\displaystyle \int\limits_{2}^{\infty} \frac{ 1 }{x } dx\)

would diverge, right?

(Because \(\displaystyle \sum_{x=2}^{\infty} \frac{ 1 }{x }\) diverges ....
either because it is a harmonic series and we should all know this, or because of the integral test (ALWAYS!! either integral and series both diverge, or both converge), or due to at lest 20 other reasons/proofs of this ....)

If \(\displaystyle \int\limits_{2}^{\infty} \frac{ 1 }{x } \) diverges, wouldn't a series that is larger diverge as well?

Answer 15

\(\displaystyle \sum_{x=2}^{\infty} \frac{ 1 }{x } < \sum_{x=2}^{\infty} \frac{ 1 }{\ln x } \)

Would you agree with this statement (right above)?

Answer 16

and if 1/x diverges, than all the more so a larger series with 1/ln(x) diverges.

(Note: Why is it larger, because ln(x) is smaller than x, so you are dividing by a smaller value, and thus getting a bigger result)

Answer 17

idku used what is known as the integral test.
http://prntscr.com/autsto
the conditions are that the function is decreasing and positive

Answer 18

actually it would be better if the website wrote
"the series and the integral both converge/diverge together".

Answer 19

yeah, that would be more clear.

Answer 20

If you understand the tests for convergence (comparison test, integral test, divergence test ....) then after some practice you will be able to determine convergence/divergence of a series (and via the integral test, you will be able to determine the convergence of an integral as well) in a snapshot.

Answer 21

For example:
\(\displaystyle \int_2^\infty\frac{1}{\ln(x!)}> \int_3^\infty\frac{1}{\ln(x^x)}=\int_3^\infty\frac{1}{x\ln(x)}= \\[1.9em] \displaystyle -\lim_{x\to\infty}\ln(\ln3))+\lim_{x\to\infty}\ln(\ln x))\to{\rm diverges}\)

Answer 22

So, what you have here is two steps to solve the problem.
First idku used a comparison test, to compare the series to a simpler series
∑ 1/ln n > ∑ 1/n
http://prntscr.com/autur4
This implies ∑ 1/ln n diverges.

Then we used an integral test, which says the integral and the series must diverge together.

If we could combine this in one step that would be ideal, but we can't integrate this function easily.

Answer 23

The fact that the limit goes to zero does not tell us much. It may or may not converge.
compare the improper integrals
\( \large \int_{1}^{\infty} \frac {1}{x^2} \)
and
\( \large \int_{1}^{\infty} \frac {1}{x} \).
These are easy to integrate so you can see two cases where the limit goes to zero, yet it both converges and diverges.
Of course if the limit does not go to zero, the area under the curve will diverge
and there is no chance it will converge.
Short proof : suppose the limit of f(x) is small nonzero constant.
infinity * small constant = infinity