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privetek

  • 2 years ago

determine whether the series converges or diverges:

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  1. privetek
    • 2 years ago
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    \[\sum_{1}^{\infty} (lnk)/(e ^{\sqrt{k}})\]

  2. EarthCitizen
    • 2 years ago
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    0

  3. privetek
    • 2 years ago
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    what test did you use?

  4. EarthCitizen
    • 2 years ago
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    when u find the limiting value \[U_{k+1}/U_{n}\]

  5. EarthCitizen
    • 2 years ago
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    \[\left| U _{k+1}/U_{k} \right|\]

  6. privetek
    • 2 years ago
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    can you kinda show me how you got zero?

  7. EarthCitizen
    • 2 years ago
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    \[ U_{k} = \ln(k)/e^{k} , U_{k+1} = \ln(k+1)/e^{k+1} \]\[ \therefore \left| U_{k+1}/U_{k} \right| = \ln(k+1)/e^{k+1} \times e^{k}/\ln(k)\]

  8. EarthCitizen
    • 2 years ago
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    \[ \ln(1) \times e^{\sqrt(k)-\sqrt(k+1)} = 0\]

  9. EarthCitizen
    • 2 years ago
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    did that help ?

  10. privetek
    • 2 years ago
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    great! thanks :)

  11. EarthCitizen
    • 2 years ago
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    what was the answer in the text book ?

  12. privetek
    • 2 years ago
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    it wasn't from my textbook so i don't know the right answer but that looks right.

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spraguer (Moderator)
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