anonymous
  • anonymous
How can I prove int(1/lnx,x=1..infinity) diverges?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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goformit100
  • goformit100
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
anonymous
  • anonymous
ok
amistre64
  • amistre64
solve it for the interval 1 to b, then take the limit as b approaches infinity

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amistre64
  • amistre64
its called an improper integral
amistre64
  • amistre64
might have to split it into 2 parts and limit it to 1 as well .... jsut an observation is all
anonymous
  • anonymous
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.
amistre64
  • amistre64
what do you get fir int (1/lnx) dx ?
anonymous
  • anonymous
I'm not trying to solve them, I just want to prove one or both of them diverges with a comparison or something
ybarrap
  • ybarrap
Compare 1/lnx to 1/x, which you know diverges. If 1/lnx > 1/x, then you're done
experimentX
  • experimentX
\[ \int_1^\infty \frac1{\log (x) }dx = \int_1^e \frac1{\log (x) }dx +\int_e^\infty \frac1{\log (x) }dx\] \[\int_e^\infty \frac1{\log (x) }dx > \int_1^\infty \frac1{x }dx\] let log(x) = z \[\int_1^e \frac1{\log (x) }dx =\int_0^1 \frac{e^z}{z }dx >\int_0^1 \frac{1}{z }dx=\int_1^\infty \frac{1}{z }\]
experimentX
  • experimentX
or simply write \[ \int_1^\infty \frac{1}{\log (x) }dx>\int_e^\infty \frac{1}{\log (x) }dx>\int_1^\infty \frac{1}{x }dx\]
anonymous
  • anonymous
How do you know x1?
experimentX
  • experimentX
log(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
experimentX
  • experimentX
woops!! for x>1, let x = x+1 log(1+x) = x - x^2/2 + x^3/3 - ... > x for 0
experimentX
  • experimentX
also do you know limit comparison test?
anonymous
  • anonymous
Just a sec
anonymous
  • anonymous
I meant "How do you know x*>*log(x) when x>1?"
experimentX
  • experimentX
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.
experimentX
  • experimentX
the max value this can reach is ln(2) - 1
experimentX
  • experimentX
log(1+x) < x - (1-ln(2)) < x
anonymous
  • anonymous
ok, i got it. Thanks
experimentX
  • experimentX
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.

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