Question

How can I prove int(1/lnx,x=1..infinity) diverges?

Answer 1

First of all @piemelke : A Warm Welcome to 'Open Study'. Please Read CoC (compulsory to read by all "Open Study" users) : http://openstudy.com/code-of-conduct

Answer 2

ok

Answer 3

solve it for the interval 1 to b, then take the limit as b approaches infinity

Answer 4

its called an improper integral

Answer 5

might have to split it into 2 parts and limit it to 1 as well .... jsut an observation is all

Answer 6

yes I know that, part. And then I'm supposed to show at least one of them diverges and that's where I'm stuck.

Answer 7

what do you get fir int (1/lnx) dx ?

Answer 8

I'm not trying to solve them, I just want to prove one or both of them diverges with a comparison or something

Answer 9

Compare 1/lnx to 1/x, which you know diverges. If 1/lnx > 1/x, then you're done

Answer 10

\[ \int_1^\infty \frac1{\log (x) }dx = \int_1^e \frac1{\log (x) }dx +\int_e^\infty \frac1{\log (x) }dx\]
\[\int_e^\infty \frac1{\log (x) }dx > \int_1^\infty \frac1{x }dx\]

let log(x) = z
\[\int_1^e \frac1{\log (x) }dx =\int_0^1 \frac{e^z}{z }dx >\int_0^1 \frac{1}{z }dx=\int_1^\infty \frac{1}{z }\]

Answer 11

or simply write
\[ \int_1^\infty \frac{1}{\log (x) }dx>\int_e^\infty \frac{1}{\log (x) }dx>\int_1^\infty \frac{1}{x }dx\]

Answer 12

How do you know x<log(x) when x>1?

Answer 13

log(x)<log(1+x) for x>0

log(1+x) = x - x^2/2 + x^3/3 - ...
you can show that the series x^2/2 - x^3/3 + ... is always positive for 0<x<1 using alternating test.

Answer 14

woops!! for x>1, let x = x+1

log(1+x) = x - x^2/2 + x^3/3 - ... > x for 0<x<1.
just by integrating 1/log(x) from 1 to 2 you can show that the integral diverges.

Answer 15

also do you know limit comparison test?

Answer 16

Just a sec

Answer 17

I meant "How do you know x*>*log(x) when x>1?"

Answer 18

let x be in between 0 and 1
you know that
x^2/2 > x^3/3 and so on.
this whole thing
-x^2/2+x^3/3 - .... is negative.

Answer 19

the max value this can reach is ln(2) - 1

Answer 20

log(1+x) < x - (1-ln(2)) < x

Answer 21

ok, i got it. Thanks

Answer 22

also this one is more trivial
log(x) = y
x = e^(y) = 1 + log(x) + log(x)^2/2! + ... the result is more trivial from here.