1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    sharks

    User Avatar
    Gold Member

    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

    sharks

    User Avatar
    Gold Member

    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

    sharks

    User Avatar
    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.
    Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook