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Determining Convergence/Divergence

  1. Feb 1, 2007 #1
    I need to determine the convergence of the following equation:
    [tex]\sum_{n=0}^\infty \frac{2^n}{3^n+5}[/tex]

    It's not necessary to be formal, but I would like an explination of how it's done. My belief is that it would converge to zero because although the limit is infinity over infinity, the [tex]3^n[/tex] trumps the [tex]2^n[/tex] . I tried L'Hopital's rule, however you just end up with [tex]\frac{\ln(2) * 2^n}{\ln(3) * 3^n}[/tex] over and over. I have not tried the integral technique but I don't believe that would work. Any suggestions? The sequence is geometric I think.
     
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  3. Feb 1, 2007 #2

    quasar987

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    Do you know what a geometric series is and what is the value of convergence of such series (i.e. does it converge or diverge?)?
     
  4. Feb 1, 2007 #3
    Yes, I know that a geometric series is [tex]a_n = a_1 * r^{n-1}[/tex], that [tex]\sum_{k=1}^n a_1 * r^{k-1} = \frac{a_1(1-r^n)}{1-r}[/tex], and that a geometric series is convergent if the absolute value of r is less than 1.
     
    Last edited: Feb 1, 2007
  5. Feb 1, 2007 #4
    Could I make that [tex]\frac{2}{3}^n[/tex] which would be divergent?
     
  6. Feb 1, 2007 #5

    quasar987

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    Ok, then consider the series

    [tex]\sum_{n=0}^\infty \frac{2^n}{3^n}[/tex]

    This is geometric with a_1=1 and r=2/3, so it converges.

    Now notice that in your series,

    [tex]\sum_{n=0}^\infty \frac{2^n}{3^n+5}[/tex]

    because of the aditional "+5" in the denominator, each term of your series is smaller than its corresponding term in the above geometric series (i.e. for each n, the n-th term of your series is smaller than the n-th term of the geometric series). So the sum of your series must be also smaller than the sum of the geometric series!

    This is the reasoning behind most convergence tests out there.


    Also note that in your first post, you arrived at the limit of [itex]\frac{\ln(2) * 2^n}{\ln(3) * 3^n}[/itex]. This is of the form [itex]ka^n[/itex] with k a constant and |a|<1, which goes to zero (ever noticed that when you square a (positive) number lesser than one, the result is always lesser than what you started with? And if you cube it, you end up with something even smaller?).

    But, and this is what I wanted to warn you about, remember that knowing that the general term goes to zero does not allow you to draw any conclusion whatsoever about the convergence of the series. It can still converge or diverge. Only when the general term does not go to zero can you say something about the convergence (namely, that the series diverge).
     
    Last edited: Feb 1, 2007
  7. Feb 1, 2007 #6
    Got it! Thank you very much! I appreciate the help.
     
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