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anonymous

  • one year ago

an=2^-n cosn pi

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  1. anonymous
    • one year ago
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    so? solve for n? or what

  2. anonymous
    • one year ago
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    determine whether the sequence converges or diverges, if it converges give the limits

  3. anonymous
    • one year ago
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    ohh

  4. anonymous
    • one year ago
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    \[a_n=\frac{\cos(n\pi)}{2^n}\]

  5. anonymous
    • one year ago
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    naw dude

  6. anonymous
    • one year ago
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    the denominator goes to zero lickety split the numerator is stuck between -1 and 1

  7. anonymous
    • one year ago
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    so the answer is 0?

  8. anonymous
    • one year ago
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    yes of course

  9. anonymous
    • one year ago
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    oh i said it wrong the denominator goes to INFINITY lickety split, not zero

  10. anonymous
    • one year ago
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    so the limit is zero

  11. anonymous
    • one year ago
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    and of course that means it converges correct?

  12. anonymous
    • one year ago
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    yes converges to zero

  13. anonymous
    • one year ago
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    thanks sooooooo much

  14. zepdrix
    • one year ago
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    If you need some way to justify it, you could use Squeeze Theorem :)\[\large\rm -1\le \cos(n \pi)\le 1\]Divide each side by 2^n,\[\large\rm -\frac{1}{2^n}\le \frac{\cos(n \pi)}{2^n}\le \frac{1}{2^n}\]And let n approach infinity on the left and right most sides of this inequality,\[\large\rm -0\le \frac{\cos(n \pi)}{2^n}\le 0\]

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