amyna
  • amyna
Detrmine monotonicity: (0.9)^n How do i solve this?
Mathematics
katieb
  • katieb
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zepdrix
  • zepdrix
(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
amyna
  • amyna
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
zepdrix
  • zepdrix
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?

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amyna
  • amyna
yes
zepdrix
  • zepdrix
\[\large\rm \frac{a_{n+1}}{a_n}=\frac{0.9^{n+1}}{0.9^n}=0.9\lt1\qquad \forall n\in\mathbb N\]So maybe we can do just this?
amyna
  • amyna
yes this seems correct! thank you!
zepdrix
  • zepdrix
sequences stuff is tricky! >.<
amyna
  • amyna
ya i know lol!

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