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anonymous

  • 5 years ago

Does this infinte series converge or diverge? Sigma from 1-infinity of 1n(n)/n^1.02

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  1. anonymous
    • 5 years ago
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    n^(1.02)?

  2. anonymous
    • 5 years ago
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    \[\sum_{1}^{\infty} \ln(n)/n^1.02\]

  3. anonymous
    • 5 years ago
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    Use the integral test you will find it will converge.

  4. anonymous
    • 5 years ago
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    Are you familiar with integral test? Or should I go on?

  5. anonymous
    • 5 years ago
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    In case you aren't If \[\int\limits_{k}^{\infty} f(x) dx \] is convergent implies \[\sum_{k}^{\infty}\an] is convergent

  6. anonymous
    • 5 years ago
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    \[\sum_{k}^{\infty}an\] is convergent

  7. anonymous
    • 5 years ago
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    take \[f(x) = \frac{\ln(n)n^(1.02)}\] and k = 1

  8. anonymous
    • 5 years ago
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    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. =]

  9. anonymous
    • 5 years ago
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    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?

  10. anonymous
    • 5 years ago
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    Yes do it by parts

  11. anonymous
    • 5 years ago
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    let u=ln(n), dv=n^(1.02)? see but from here what would v be?

  12. anonymous
    • 5 years ago
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    sorry, i don't mean to hijack the problem, it's just the same thing i've been having trouble with, too.

  13. anonymous
    • 5 years ago
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    oh and i meant dv=n^(-1.02)dx

  14. anonymous
    • 5 years ago
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    unless i'm setting u and dv to the wrong parts

  15. anonymous
    • 5 years ago
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    v is something stupid -50/n^(0.02)

  16. anonymous
    • 5 years ago
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    ...what. where did the 50 come in?

  17. anonymous
    • 5 years ago
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    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?

  18. anonymous
    • 5 years ago
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    1.02 = 1 + 1/50, sorry I hate decimals

  19. anonymous
    • 5 years ago
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    oh right, i see. thanks.

  20. anonymous
    • 5 years ago
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    Are we good?

  21. anonymous
    • 5 years ago
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    oh i am so good with this now. but then i definitely stole this question from the original poster, so... sorry about that, Aubun.

  22. anonymous
    • 5 years ago
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    Okay! I'm off to bed, have a big math test tomorrow

  23. anonymous
    • 5 years ago
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    good luck!

  24. anonymous
    • 5 years ago
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    Thank You

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