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amyna

  • one year ago

Detrmine monotonicity: (0.9)^n How do i solve this?

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  1. zepdrix
    • one year ago
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    (0.9)^1 = 0.9 (0.9)^2 = 0.81 (0.9)^3 = 0.729 It's clearly monotonically decreasing as every term is smaller than the one before it. But how to show this -_- Hmm this calc 2? I don't remember the formulas lol

  2. amyna
    • one year ago
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    yes its calc 2, i know its decreasing, but i have to prove it. My teacher said there are 2 ways: 1) is to take the an+1 sequence thing 2) is to take the derivative

  3. zepdrix
    • one year ago
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    If the sequence is decreasing, then the ratio\[\large\rm \frac{a_{n+1}}{a_n}\lt 1.\]Because the numerator should be `smaller` if it's decreasing, ya?

  4. amyna
    • one year ago
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    yes

  5. zepdrix
    • one year ago
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    \[\large\rm \frac{a_{n+1}}{a_n}=\frac{0.9^{n+1}}{0.9^n}=0.9\lt1\qquad \forall n\in\mathbb N\]So maybe we can do just this?

  6. amyna
    • one year ago
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    yes this seems correct! thank you!

  7. zepdrix
    • one year ago
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    sequences stuff is tricky! >.<

  8. amyna
    • one year ago
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    ya i know lol!

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