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Hitaro9

  • one year ago

Does the Series ke^-k^2 converge or diverge?

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  1. hitaro9
    • one year ago
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    \[\sum_{k=1}^{infinity}k e^{-k ^{2}}\]

  2. anonymous
    • one year ago
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    $$\large \sum_{k=1}^{\infty}k e^{-k ^{2}}= $$ notice that is close to the derivative of \( \large e^{-x^2} \)

  3. hitaro9
    • one year ago
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    Oh, right, there's a thing in the book about an integral test. We haven't discussed it in class yet, but, I'm guessing that's what this is?

  4. anonymous
    • one year ago
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    sort of, I was thinking of this http://prntscr.com/8j57eg

  5. hitaro9
    • one year ago
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    That looks significantly more complicated than what we've been doing thus far.

  6. anonymous
    • one year ago
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    but now that you mention it, you can use integral test

  7. hitaro9
    • one year ago
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    Okay then, that's good. I'm reading up on it now, might ask some questions

  8. IrishBoy123
    • one year ago
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    applied the ratio test yet?

  9. hitaro9
    • one year ago
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    Ratio test didn't seem to work, unless I made some algebraic errors

  10. IrishBoy123
    • one year ago
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    \[\large \frac{(k+1) e^{-(k+1) ^{2} }}{k e^{-k ^{2}}}=\]

  11. hitaro9
    • one year ago
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    Right, then I ended up with some mess with the exponents that didn't seem to work out nicely

  12. IrishBoy123
    • one year ago
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    \[\large \frac{k+1}{k} e^{-(2k+1)}\]

  13. hitaro9
    • one year ago
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    To save you the work of typing out a whole bunch of equations, if there's some way to do it with the ratios, I should be able to handle it if you can help me through the algebra

  14. IrishBoy123
    • one year ago
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    right so far?

  15. hitaro9
    • one year ago
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    Umm one moment. That's not what I did so that's probably where I went wrong

  16. IrishBoy123
    • one year ago
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    expand the top exponent and subtract the bottom one

  17. hitaro9
    • one year ago
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    Right that makes sense. I dunno why I didn't see that.

  18. anonymous
    • one year ago
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    because the e^(-2k+1) is in the denominator. nice work irishboy

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