anonymous
  • anonymous
Does this infinte series converge or diverge? Sigma from 1-infinity of 1n(n)/n^1.02
Mathematics
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anonymous
  • anonymous
Does this infinte series converge or diverge? Sigma from 1-infinity of 1n(n)/n^1.02
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
n^(1.02)?
anonymous
  • anonymous
\[\sum_{1}^{\infty} \ln(n)/n^1.02\]
anonymous
  • anonymous
Use the integral test you will find it will converge.

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anonymous
  • anonymous
Are you familiar with integral test? Or should I go on?
anonymous
  • anonymous
In case you aren't If \[\int\limits_{k}^{\infty} f(x) dx \] is convergent implies \[\sum_{k}^{\infty}\an] is convergent
anonymous
  • anonymous
\[\sum_{k}^{\infty}an\] is convergent
anonymous
  • anonymous
take \[f(x) = \frac{\ln(n)n^(1.02)}\] and k = 1
anonymous
  • anonymous
f(x) = ln(n)/n^(1.02) take the integral from 1 to inf and show it does not diverge that is it has a finite answer. That will mean our series also converges. =]
anonymous
  • anonymous
i am also interested in this problem, and i'd love to use the integral test, but i don't know how to integrate that. would you do it by parts?
anonymous
  • anonymous
Yes do it by parts
anonymous
  • anonymous
let u=ln(n), dv=n^(1.02)? see but from here what would v be?
anonymous
  • anonymous
sorry, i don't mean to hijack the problem, it's just the same thing i've been having trouble with, too.
anonymous
  • anonymous
oh and i meant dv=n^(-1.02)dx
anonymous
  • anonymous
unless i'm setting u and dv to the wrong parts
anonymous
  • anonymous
v is something stupid -50/n^(0.02)
anonymous
  • anonymous
...what. where did the 50 come in?
anonymous
  • anonymous
i can see the power rule making the exponent come down to 0.02, and also kicking the negative out front, but 50? where did that come from?
anonymous
  • anonymous
1.02 = 1 + 1/50, sorry I hate decimals
anonymous
  • anonymous
oh right, i see. thanks.
anonymous
  • anonymous
Are we good?
anonymous
  • anonymous
oh i am so good with this now. but then i definitely stole this question from the original poster, so... sorry about that, Aubun.
anonymous
  • anonymous
Okay! I'm off to bed, have a big math test tomorrow
anonymous
  • anonymous
good luck!
anonymous
  • anonymous
Thank You

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