Mathematics
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OpenStudy (anonymous):
Σ 1/(nlnn) from infi to n=2
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OpenStudy (anonymous):
it the question "does it converge"?
OpenStudy (freckles):
from n=2 to inf ?
OpenStudy (anonymous):
because it seem unlikely
OpenStudy (anonymous):
The question is does this convergent or divergent
OpenStudy (anonymous):
try \[\int_2^{\infty}\frac{dx}{x\ln(x)}\]
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OpenStudy (anonymous):
Yes n=2 to infi
OpenStudy (anonymous):
use the integral test
OpenStudy (anonymous):
and the integral is done by a u - sub \(u=\ln(x)\)
OpenStudy (anonymous):
Du = 1/x dx
OpenStudy (anonymous):
yup, right in front of you, just what you need
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OpenStudy (anonymous):
So 1/u du
OpenStudy (anonymous):
hmmm yup
OpenStudy (anonymous):
Is this hormonic series?
OpenStudy (anonymous):
it is an integral not a series
OpenStudy (anonymous):
Oh
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OpenStudy (anonymous):
and no, harmonic series is \(\sum \frac{1}{n}\)
compute the anti - derivative and see what you get
OpenStudy (anonymous):
\[\int \frac{du}{u}=?\]
OpenStudy (anonymous):
Hm let me see
OpenStudy (anonymous):
Ln u
OpenStudy (anonymous):
k right
and \(u=\ln(x)\) so final answer?
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OpenStudy (anonymous):
Ln(lnx)?
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
final job it so compute
\[\lim_{x\to \infty}\ln(\ln(x))\]
OpenStudy (anonymous):
Then how am I define if it con or div
OpenStudy (anonymous):
Oh do limit
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OpenStudy (anonymous):
Ln(ln(infi)) =infi
OpenStudy (anonymous):
yes
so the integral diverges, and therefore so does the series
OpenStudy (anonymous):
Ah I see so it div
OpenStudy (anonymous):
Thank you so much sir