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Series test ?

  1. May 23, 2012 #1
    [tex] \sum_{r=1}^{\infty} \frac{r!}{3^{r^{2}}} [/tex]

    My solution:

    [tex] \frac{3^{r^2}}{r!} > r^2 [/tex]

    So [tex] \frac{r!}{3^{r^2}} < \frac{1}{r^2} [/tex]

    So as 1/r^2 converges, it converges by comparison test.

    This was in my exam today, I messed up a lot leading up to it. But the question said I could use any test in general to work this out.

    It was work 6 marks which is quite a bit so is my solution okay ?
  2. jcsd
  3. May 23, 2012 #2


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    Use the ratio test. The latter is always preferable whenever factorials are involved.
  4. May 23, 2012 #3
    It was in my exam, it's already over. I did the ratio test for some reason cut it out and put this instead as it was smaller.

    It's correct right ?
  5. May 23, 2012 #4


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    How do you know this?? I'm not saying it's wrong, but it should have an explanation.
  6. May 23, 2012 #5
    I evaluated a few terms during the exams(beginning with 1) its not a little bigger it's ALOOOT bigger. So much so that you can't evaluate beyond r=13 as size of the number is too big.

    At r=14 there are more digits in the result than atoms in the universe so it kinda goes without saying that its bigger than r^2.

    Regardless, I'm hoping no marks are deducted as we are interested in the asymptotic difference and if it's huge...
    Last edited: May 23, 2012
  7. May 23, 2012 #6


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    Here's my attempt using ratio test:
    [tex]L= \lim_{r\to\infty}\frac{u_{n+1}}{u_n}=\lim_{r\to \infty} \left( \frac{(r+1)r!}{3^{r^2}.3^{2r}.3}\times \frac{3^{r^{2}}}{r!} \right) =\lim_{r\to\infty}\frac{(r+1)}{3^{2r}.3}= \lim_{r\to\infty}\frac {(r+1)}{9^r.3}[/tex]The limit is an indeterminate form ∞/∞, so using L'Hopital's rule:[tex]\lim_{r\to\infty}\frac {1}{9^r.3\ln 9}=\frac{1}{\infty}=0[/tex]
    Since L < 1, therefore the original series converges.
    Last edited: May 23, 2012
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