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Homework Help: Convergence/Divergence of a Series

  1. Jul 11, 2008 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution
    I decided to use the ratio test:

    [tex]\frac{1}{2^{n+1}+(\frac{1}{3})^{n+1}}[/tex] x [tex]\frac{2^{n}+(\frac{1}{3})^{n}}{1}[/tex]

    And I got [tex]lim_{n\rightarrow\infty}[/tex][tex]\frac{1}{2 + \frac{1}{3}}[/tex]

    But I'm A) pretty sure it's wrong and B) if not, what do I do after that step?
  2. jcsd
  3. Jul 11, 2008 #2


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    Homework Helper

    You can't cancel the term [tex]2^n + (\frac{1}{3})^n[/tex] just like that since [tex]2^{n+1} + (\frac{1}{3})^{n+1} \ \mbox{is not} \ (2^n + (\frac{1}{3})^n) \cdot (2 + \frac{1}{3})[/tex]

    Instead express [tex]2^n + (\frac{1}{3})^n[/tex] as a fraction and do the same for the (n+1) expression as well. Something will cancel out and then you should be able to apply a certain limit rule to get the answer.
  4. Jul 11, 2008 #3
    I thought I could write 3n+1 as (3)(3n). So I should combine the two so that 2n + [tex]\frac{1}{3^{n}}[/tex] = [tex]\frac{7}{6}[/tex][tex]^{n}[/tex] and then find its limit?
  5. Jul 11, 2008 #4
    OH! Actually, could I write the f(x) as a fraction so I would get [tex]\frac{1}{\frac{7}{6}^{n}}[/tex] and then find its convergence/divergence through a geometric series? Or should I use an integral test?
  6. Jul 11, 2008 #5


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    Homework Helper

    3^(n+1) = 3(3^n). But that isn't what you're doing here. Just express [tex]2^{n+1} + \frac{1}{3^{n+1}}[/tex] as a single fraction. Then you can apply the ratio test easily.
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