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Is this series convergent or divergent.

  1. Dec 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Ʃ ne(-n2)

    2. Relevant equations

    3. The attempt at a solution

    I used the ratio test and wanted to know if the way I did it is correct or not

    |a(n+1) / a(n)|

    n+1 (e(-n2 -2n-1)) / n e(-n2)

    Now e-n^2 cancels and we get

    limn→∞ n+1/n * 1/(e2n)(e)

    After you take the limits you get (1)*0 = 0 < 1 so it's convergent
  2. jcsd
  3. Dec 14, 2013 #2


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    I have added your missing brackets. Aside from that your working is correct.
  4. Dec 14, 2013 #3


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    Sure, this works. But this sum looks almost purpose-built for the integral test. Try it (if you're allowed to use it).
  5. Dec 14, 2013 #4
    I am actually but I just wanted to see if the ratio test worked

    Thanks for the replies!
  6. Dec 14, 2013 #5

    Ray Vickson

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    You really need to learn how to write in ASCII---in particular, you need to use brackets. Your first equation AS WRITTEN says
    [tex] \left|\frac{a_{n+1}}{a_n}\right| = n +1 \frac{e^{-n^2-2n-1}}{n} e^{-n^2}.[/tex]
    Your second formula reads as
    [tex] \lim_{n \to \infty} n + \frac{1}{n} \frac{1}{e^{2n}} e [/tex]
    I hope these are not what you mean. I hope you intended the first one to be
    [tex] \left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)e^{-n^2-2n-1}}{n e^{-n^2}},[/tex]
    etc. To make sure this happens you need parentheses!
  7. Dec 15, 2013 #6
    Yeah I'm not very good with ASCII I'm still learning how to use them.
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