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Determining Series Convergence using the Ratio Test

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data

    I'm asked to specifically use the Ratio Test (formula below) to determine whether this series converges or diverges (if it converges, the value to which it converges is not needed.)

    [tex]\sum_{n=1}^{\infty}\frac{n}{(e^n)^2}[/tex]

    2. Relevant equations

    Ratio Test:
    If [itex]a_n[/itex] is a sequence of positive terms, then the following conditions are true for the limit:
    [tex]\lim_{n\to\infty}\frac{a_{n+1}}{a_n}[/tex]
    If lim<1, the series converges.
    If lim>1, the series diverges.
    If lim=1, inconclusive.

    3. The attempt at a solution

    By simply plugging values into the formula and rearranging the resulting fractions, I get this form:

    [tex]\frac{(e^n)^2(n+1)}{(e^{(n+1)})^2\cdot n}[/tex]

    From what I can see, applying L'Hopital's rule won't help get this limit into a solvable form, and no algebraic transformations pop out at me as being helpful.

    How would one start solving this limit?
     
    Last edited: Sep 26, 2011
  2. jcsd
  3. Sep 26, 2011 #2
    [tex]
    \frac{e^{n^2}}{e^{(n + 1)^2}} = e^{-(n + 1)^2 + n^2} = e^{-2n - 1}
    [/tex]
     
  4. Sep 26, 2011 #3
    And to make that even more explicit

    [tex]e^{-(2n + 1)}[/tex]
     
  5. Sep 26, 2011 #4
    Hmm.. I tried that method, but the limit evaluates to 0, which doesn't help when identifying convergence.

    I could try a bunch of other tests on this series, but the assignment specifically says to use the Ratio Test. Any suggestions? Perhaps I could just say that I performed the Ratio Test, but it was inconclusive, so I used another method.
     
  6. Sep 26, 2011 #5
    ORLY? Care to tell us what ratio test you are using?
     
  7. Sep 26, 2011 #6
    I just factored out the term provided above, and got:

    [tex]e^{-(2n+1)}\cdot(\frac{n+1}{n})[/tex]
    [tex]\frac{1+\frac{1}{n}}{e^{(2n+1)}}[/tex]

    And that approaches 0 as [itex]n\to\infty[/itex]: [itex]\frac{1}{\infty}[/itex].
     
  8. Sep 26, 2011 #7
    So, how is this inconclusive?
     
  9. Sep 26, 2011 #8
    Because according to the Ratio Test, if the limit in this form equates to 0, then I can't say anything regarding the convergence of the series.
     
  10. Sep 26, 2011 #9
    And when could you say that it converges or diverges?
     
  11. Sep 26, 2011 #10
    I wrote the conditions of the Ratio Test in the OP. If the limit is greater than 1, we can say that it diverges. If it's less than 1, the series converges.
     
  12. Sep 26, 2011 #11
    So, is zero greater than or smaller than one?
     
  13. Sep 26, 2011 #12
    I think you mean "if the limit in this form equates to 1" then it's inconclusive.
     
  14. Sep 26, 2011 #13
    OOPS! I memorized the theorem wrong! If it EQUALS 1, the test is inconclusive. Never mind.

    Sometimes I wonder how I'm still alive after these retarded mistakes.

    Thanks! You're a life saver :)

    Edit:


    Yeah, I figured that out a little too late. :P Thanks :)
     
    Last edited: Sep 26, 2011
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