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anonymous

  • one year ago

Find the values of x so that the series below converges. (((x+8)^(n))/(2^n)) please help find the interval. goes to INF and n=0

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  1. thomas5267
    • one year ago
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    Is this the equation? \[ a_n=\frac{(x+8)^n}{2^n} \]

  2. anonymous
    • one year ago
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    yes

  3. thomas5267
    • one year ago
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    What is the required limit?

  4. anonymous
    • one year ago
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    I'm looking to find the values of x so that the series below converges. (the interval it does

  5. anonymous
    • one year ago
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    limit is n=0

  6. anonymous
    • one year ago
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    to INF

  7. anonymous
    • one year ago
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    does that makes sense can u help?

  8. thomas5267
    • one year ago
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    So the question is find x such that: \[ \lim_{n\to0}\frac{(x+8)^n}{2^n}=\infty \]

  9. anonymous
    • one year ago
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    no it has the simga logo in front of the fraction with a INF aboe th sigma and a n=0 below

  10. anonymous
    • one year ago
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    \[\sum_{n=0}^{INF}\]

  11. thomas5267
    • one year ago
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    Find such x such that: \[ \sum_{n=0}^\infty\frac{(x+8)^n}{2^n}=\infty \]

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