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Determine Series' Convergence or Divergence

  1. May 7, 2012 #1
    1. The problem statement, all variables and given/known data
    ##\sum _{n=1}ne^{-n}##

    2. Relevant equations
    Ratio Test
    Integral Test

    3. The attempt at a solution
    I know that by the ratio test, it converges absolutely. But, I am unable to determine its convergence through the integral test . Could someone help? I thought that the integral test would be best, since e is used.

    The ratio test gives me 1/e approx 0.367879 <1; therefore, it converges absolutely.

    Now, the integral test is my concern. I believe I have done all correct steps, but the result shows that it diverges.
    ##\lim _{b\rightarrow \infty }\int _{1}^{b}\dfrac {n} {e^{n}}dn##

    Integrating by parts with: u=n; du=dn; dv=(e^-n) dn; v=-e^-n
    ## \left[ -e^{-n}\left( n\right) +-e^{-n}\right] _{1}^{\infty }##
    The final result is:
    ##-\dfrac {\infty } {e^{\infty} }-\dfrac {1} {e^{\infty }}+\dfrac {2} {e} ## and it diverges. What am I doing wrong?
    Thank you.
  2. jcsd
  3. May 7, 2012 #2
    What is ##\displaystyle \lim_{n \rightarrow \infty} \frac{n}{e^n}##? You will need to use l'Hospital's rule!
  4. May 7, 2012 #3
    Wow. Thank you.
  5. May 7, 2012 #4
    My integral after L'Hospital doesn't agree with answers online. Could you tell me what's wrong?
    ##\lim _{b\rightarrow \infty }\int _{1}^{b}\dfrac {n} {e^{n}}dn##

    ##\int _{1}^{b}\dfrac {1} {en}dn=\left[ -e^{-n}\right] _{1}^{b}=-\dfrac {1} {e^{\infty }}+\dfrac {1} {e}##
    , the first term to the left is e^n
  6. May 7, 2012 #5


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    You need to find the limit of [itex]\frac{1}{e^n}[/itex] and not its integral.
  7. May 7, 2012 #6
    The limit of (1/ e^n) as n approaches infinity is 0. The test for divergence is inconclusive.

    I'm trying to find the determination of convergence or divergence through the integral test.
    Last edited: May 7, 2012
  8. May 7, 2012 #7


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    OK, i've worked it out, and the limit should be: [itex]\frac{2}{e}[/itex]
    Last edited: May 7, 2012
  9. May 7, 2012 #8
    Yeah, that's what it's shown online.
    I get 1/e. My procedure is above. Could you help?
  10. May 7, 2012 #9


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    Gold Member

    This step is correct:
    [tex]\left[ -e^{-n}\left( n\right) -e^{-n}\right] _{1}^{\infty }[/tex]
    Let b=∞, then,
    [tex]\left[ -e^{-n}\left( n\right) -e^{-n}\right] _{1}^{b}[/tex]
    This gives:
    [tex]\left[ \left( \frac{-b}{e^b} -e^{-b}\right) - \left( \frac{-1}{e} -e^{-1}\right) \right][/tex]
    Now, what you have to find is the limit as b→∞
    [tex]\lim_{b→∞} \left[ \left( \frac{-b}{e^b} -e^{-b}\right) - \left( \frac{-1}{e} -e^{-1}\right) \right][/tex]
    But the limit of the following requires using L'Hopital's rule,
    [tex]\lim_{b→∞} \frac{-b}{e^b} = \lim_{b→∞} \frac{-1}{e^b}[/tex]
    You can then work out the rest:
    [tex]\lim_{b→∞} \left[ \left( \frac{-1}{e^b} - \frac{1}{e^b}\right) - \left( \frac{-1}{e} - \frac{1}{e}\right) \right][/tex]
  11. May 7, 2012 #10
    Thanks, it's what I suspected. This L'Hospital rule keeps coming and coming.
  12. May 7, 2012 #11
    You had the answer in your first post! The first term goes to zero by l'Hospital's, the second goes to zero, and the final term is just a constant which was ##\frac{2}{e}##.
  13. May 7, 2012 #12
    I know, I didn't know I could use L'Hospital in the middle of an integration by parts.
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