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anonymous

  • 5 years ago

Power Series: sigma n=1 n-> infinity ((n(n+1)^n)/4^n) so far ive gotten IxI < (1/4) how do i find the interval of convergence and radius of convergence?

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  1. anonymous
    • 5 years ago
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    you dont have x in your summation, how can you get |x| < 1/4?

  2. anonymous
    • 5 years ago
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    woops sigma n=1 n-> infinity ((n(x+1)^n)/4^n) IxI < (1/4)

  3. anonymous
    • 5 years ago
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    hello again btw :]

  4. anonymous
    • 5 years ago
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    great :)

  5. anonymous
    • 5 years ago
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    by using the root test, you will get the limit (n^(1/n))[(x+1)/4]

  6. anonymous
    • 5 years ago
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    since lim (n^(1/n)) = 1, for the series to converge, you get |x+1| < 4

  7. anonymous
    • 5 years ago
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    oh okay hmm okay now how would I find the intervals of convergence?

  8. anonymous
    • 5 years ago
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    the radius of convergence is 4, and it is around -1. So the interval we are interested is (-5,3)

  9. anonymous
    • 5 years ago
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    okay so we plug in what makes it zero which in this case is -1, and how did you get the interval?

  10. anonymous
    • 5 years ago
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    by solving |x+1| < 4. -4 < x+1 < 4 -5 < x < 3

  11. anonymous
    • 5 years ago
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    Ohhh! xD thanks

  12. anonymous
    • 5 years ago
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    So check the convergence of the series at the end points

  13. anonymous
    • 5 years ago
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    you're welcome

  14. anonymous
    • 5 years ago
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    how would I know to apply the root test? ratio?

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