Hitaro9
  • Hitaro9
Does the Series ke^-k^2 converge or diverge?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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Hitaro9
  • Hitaro9
\[\sum_{k=1}^{infinity}k e^{-k ^{2}}\]
anonymous
  • anonymous
$$\large \sum_{k=1}^{\infty}k e^{-k ^{2}}= $$ notice that is close to the derivative of \( \large e^{-x^2} \)
Hitaro9
  • Hitaro9
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?

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anonymous
  • anonymous
sort of, I was thinking of this http://prntscr.com/8j57eg
Hitaro9
  • Hitaro9
That looks significantly more complicated than what we've been doing thus far.
anonymous
  • anonymous
but now that you mention it, you can use integral test
Hitaro9
  • Hitaro9
Okay then, that's good. I'm reading up on it now, might ask some questions
IrishBoy123
  • IrishBoy123
applied the ratio test yet?
Hitaro9
  • Hitaro9
Ratio test didn't seem to work, unless I made some algebraic errors
IrishBoy123
  • IrishBoy123
\[\large \frac{(k+1) e^{-(k+1) ^{2} }}{k e^{-k ^{2}}}=\]
Hitaro9
  • Hitaro9
Right, then I ended up with some mess with the exponents that didn't seem to work out nicely
IrishBoy123
  • IrishBoy123
\[\large \frac{k+1}{k} e^{-(2k+1)}\]
Hitaro9
  • Hitaro9
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
IrishBoy123
  • IrishBoy123
right so far?
Hitaro9
  • Hitaro9
Umm one moment. That's not what I did so that's probably where I went wrong
IrishBoy123
  • IrishBoy123
expand the top exponent and subtract the bottom one
Hitaro9
  • Hitaro9
Right that makes sense. I dunno why I didn't see that.
anonymous
  • anonymous
because the e^(-2k+1) is in the denominator. nice work irishboy

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