# Strictly decreasing sequence (analysis)

1. Dec 22, 2009

### riskandar

1. The problem statement, all variables and given/known data

f(x) = (3 sin(x))/(2 + cos(x))

let x_0 be a number in (0, (2/3)*pi], and define a sequence recursively by setting
x_n+1 = f(x_n)

(1) prove that the sequence {x_n} is strictly decreasing sequence in (0, (2/3)*pi] and that lim x_n =0

(2) Find an integer k greater or equal to 1 such at {n^(1/k) x_n} is convergent to a finite, non-zero real number and evaluate lim n^(1/k) x_n

2. Relevant equations

3. The attempt at a solution

I tried the first problem by induction but I don't know how to prove it for P(n+1) (assuming P(n) is true). Any help is appreciated thank you.

2. Dec 22, 2009

### ystael

Don't think about this as an inductive argument at the bottom, but as an inequality.

Because your sequence is defined by iterating the function $$f$$, the simplest way to prove that $$x_{n+1} = f(x_n) < x_n$$ for all $$n$$ is to prove that $$f(x) < x$$ for all $$0 < x \leq \frac{2\pi}3$$.

(You also need to show that $$f(x)$$ still lies in $$\left(0, \frac{2\pi}3\right]$$; otherwise you can't conclude $$f(x_{n+1}) < x_{n+1}$$, because you don't know that $$x_{n+1}$$ is in the domain to which your inequality applies. This part of the argument is an induction.)

To show that $$x_n \to 0$$, you need a little more than that. When you prove the fact above, try to show not just that $$f(x) < x$$, but that $$f(x) < x$$ in some kind of systematic or predictable way. Think about examples of simple sequences that converge to zero, and try to establish a comparison with one of those.

Last edited: Dec 22, 2009