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Math Help - Another Series problem

  1. #1
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    Question Another Series problem

    Hello, I have the problem:

    I get:
    \lim_{n \to \infty} \left|\frac{(n+1)^9(x+2)^{n+1}}{9^{n+1}(n+1)^{29/3}} \cdot \frac{(9^n)(n^{29/3})}{n^9(x+2)^n} \right|
    then:
    \left |x+2 \right|\lim_{n \to \infty} \frac{(n+1)^9(n^{29/3})}{9(n+1)^{29/3}n^9}
    and I get stuck after that, is it correct so far?
    Thanks,
    Matt
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  2. #2
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    \left |x+2 \right|\lim_{n \to \infty} \frac{{\color{red}(n+1)^9}{\color{blue}(n^{29/3})}}{9{\color{red}(n+1)^{29/3}}{\color{blue}n^9}} = \frac{|x+2|}{9} \lim_{n \to \infty} \frac{n^{2/3}}{(n+1)^{2/3}} (Subtracted exponents)

    So by the ratio test, the series absolutely converges and thus converges if \frac{|x+2|}{9} < 1 \iff |x+2| < 9 \iff \hdots
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  3. #3
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    Thanks for the fast reply. Okay, I know that you took out the 9 and put it on the outside and I know when you have something like \frac{x^7}{x^5}=x^2 but could you explain a little more how you got the new fraction in the second part where you have (subtracted exponents)?
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  4. #4
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    I factored out the 1/9 and simplified the exponents (it's colour-coded!): \frac{29}{3} -3 = \frac{29}{3} - \frac{27}{3} = \frac{2}{3}. The limit of the n-containing term is 1 so .. etc. etc.
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  5. #5
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    Oh, I see... I should of saw that earlier...
    Thank you!
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  6. #6
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    Hmm, webwork is saying that my answer isn't correct, I have -7 \leq x<11 Is that correct?
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    |x+2| < 9 \ \iff \ - 9 < x + 2 < 9. You're subtracting 2 from all 3 sides of the inequality, not adding.
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  8. #8
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    So what should the interval of convergence be? I don't understand exactly what you are saying.
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  9. #9
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    -9 < x + 2 < 9 \ \iff \ -11 < x < 7 is what I'm saying. Test for convergence for x = -11, 7 and you'll find your interval of convergence.
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  10. #10
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    For -11 I got infinity and 7 came out to negative infinity when put in the series. So both do not converge. But I still can't get the interval of convergence.
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