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Help with infinite series!

  1. Feb 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the radius of convergence and interval of convergence for the following infinite series
    [tex] \sum_{n=1}^{\∞} \frac{x^n n^2}{3 \cdot 6 \cdot 9 \cdot ... (3n)} [/tex]



    2. Relevant equations
    Ratio test


    3. The attempt at a solution
    Using ratio test we get
    im not sure how to put absolute value signs but
    [tex] \frac{(n+1)^2 x^{n+1}}{3 \cdot 6 \cdot 9 ... (3n) \cdot 3(n+1)} \frac{3 \cdot 6 \cdot 9 \cdot 3n}{n^2 x^n} [/tex]

    and this becomes
    [tex] \frac{(n+1)^2 x}{3 n^2 (n+1)} [/tex]
    and that simplifies to

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

    now here is where I have the trouble. the bottom of the fraction above is 'stronger' than the top which means that when we put the above < 1, it does not solve.

    Can you please check if I did all the math correctly? Your assistance is greatly appreciated!
     
  2. jcsd
  3. Feb 2, 2013 #2

    jbunniii

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    Yes, this looks right. Now ignore the ##x## for a moment, and consider this limit:
    $$\lim_{n \rightarrow \infty} \frac{n+1}{3n^2}$$
    What does this equal?
     
  4. Feb 2, 2013 #3

    LCKurtz

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    You should have absolute values:$$
    \lim_{n \rightarrow \infty}\left| \frac {x(n+1)}{3n^2}\right |$$
    What do you get for that limit and what does it tell you?
     
  5. Feb 2, 2013 #4
    thanks for your reply.

    Isn't the limit zero? Wouldnt that mean that we can't make the limit <1?
     
  6. Feb 2, 2013 #5

    LCKurtz

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    In my book, 0 < 1. What does that tell you about the series?
     
  7. Feb 2, 2013 #6
    Ok that's perfect. that tells me that series converges.
    How would I find the radius of convergence, though?
     
  8. Feb 2, 2013 #7

    LCKurtz

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    Well, for what values of x does it converge?
     
  9. Feb 2, 2013 #8
    it would converge for all x?

    Would that mean the interval of convergence is -∞ to +∞ ?
    Does that mean the radius is infinity?
     
  10. Feb 2, 2013 #9

    LCKurtz

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    Since the limit ratio was zero no matter what value x has, the answer to all three questions is yes.
     
  11. Feb 2, 2013 #10
    Thank you for your help
     
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