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(Limit) Comparison Test

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\sum[/tex][tex]^{\infty}_{n=1}[/tex] [tex]\frac{e^{n}+n}{e^{2n}-n^{2}}[/tex]

    2. Relevant equations

    I have to use either the Comparison Test or the Limit Comparison Test to show whether the series converges or diverges.

    3. The attempt at a solution

    [tex]a_{n}[/tex] = [tex]\frac{e^{n}+n}{e^{2n}-n^{2}}[/tex]

    [tex]b_{n}[/tex] = [tex]\frac{1}{e^{2n}}[/tex]

    [tex]lim_{n->\infty}[/tex] [tex]\frac{e^{n}+n}{e^{2n}-n^{2}}[/tex] * [tex]e^{2n}[/tex]

    Annnd I'm not sure what to do beyond this point. I'm not even sure I'm taking the right equation for b[tex]_{n}[/tex]... Is it okay to just ignore the [tex]e^{n}[/tex] in the numerator like that?
     
  2. jcsd
  3. Sep 16, 2009 #2

    Office_Shredder

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    You can do the limit formally like so

    [tex] \lim_{n \rightarrow \infty} \frac{e^{3n} + e^{n}n}{e^{2n}-{n^2}}[/tex]

    Divide the top and bottom by e2n

    [tex]\lim_{n \rightarrow \infty} \frac{e^n+\frac{n}{e^n}}{1-\frac{n^2}{e^{2n}}}[/tex]

    The numerator goes to infinity as n gets large, but the denominator goes to 1. So obviously the limit doesn't exist. Since the series for bn converges, you need to find something better.

    The n and n2 in the limit don't grow nearly as fast as the exponentials, so when considering limit behavior, you might want to assume they're 0 to get a rough idea of what is going on. What can you divide [tex] \frac{e^n}{e^{2n}}[/tex] by to get that the limit as n goes to infinity still exists that might be a suitable candidate?
     
  4. Sep 16, 2009 #3
    Hmm. Could I make it [tex]\frac{e^{n}}{ne^{2n}}[/tex]? Or [tex]\frac{ln(e^{n})}{ln(e^{2n})}[/tex]?
     
  5. Sep 16, 2009 #4

    zcd

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    Try the limit comparison test with [tex]\sum_{n=1}^{\infty}\frac{1}{e^{n}-n}[/tex]
    (this converges if you compare it with the geometric series by the way)
     
  6. Sep 17, 2009 #5
    Thank you so much, everyone! zcd, is the [tex]b_{n}[/tex] you gave me less than [tex]\frac{1}{e^{n}}[/tex] though? It seems like it should be bigger because the denominator is less...
     
  7. Sep 17, 2009 #6

    Dick

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    It is bigger. But it's less than 1/((e^n)/2), for example, since (e^n-n)>((e^n)/2) if n is large enough.
     
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