Question

Detrmine monotonicity:
(0.9)^n

How do i solve this?

Answer 1

(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

Answer 2

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

Answer 3

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?

Answer 4

yes

Answer 5

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

Answer 6

yes this seems correct! thank you!

Answer 7

sequences stuff is tricky! >.<

Answer 8

ya i know lol!