# Strictly decreasing sequence (analysis)

• riskandar
In summary, The given function f(x) can be iterated to create a sequence {x_n} that is strictly decreasing in the interval (0, (2/3)*pi]. Additionally, it can be proven that the limit of this sequence is 0. To find an integer k greater or equal to 1 such that {n^(1/k) x_n} converges to a finite, non-zero real number, the function f(x) must be shown to be consistently smaller than x in the given interval. Therefore, the convergence of {n^(1/k) x_n} can be compared to known examples of sequences that converge to 0.
riskandar

## Homework Statement

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

## 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.

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.

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## 1. What is a strictly decreasing sequence?

A strictly decreasing sequence is a sequence of numbers in which each term is smaller than the previous term. In other words, the value of each term in the sequence decreases as the sequence progresses.

## 2. How is a strictly decreasing sequence different from a non-decreasing sequence?

A strictly decreasing sequence is the opposite of a non-decreasing sequence. In a non-decreasing sequence, the value of each term either stays the same or increases as the sequence progresses, while in a strictly decreasing sequence, the value of each term decreases.

## 3. Can a strictly decreasing sequence contain repeating terms?

Yes, a strictly decreasing sequence can contain repeating terms. For example, the sequence 10, 9, 9, 8, 7, 6 is a strictly decreasing sequence even though the term 9 appears twice.

## 4. What is the importance of studying strictly decreasing sequences in analysis?

Strictly decreasing sequences are important in analysis because they are used to prove the monotone convergence theorem, which states that a bounded and monotonic sequence (either strictly increasing or strictly decreasing) is convergent. This theorem is useful in many areas of mathematics, including calculus and real analysis.

## 5. How can we determine if a sequence is strictly decreasing?

To determine if a sequence is strictly decreasing, we need to compare each term to the previous term. If each term is smaller than the previous term, then the sequence is strictly decreasing. However, if there is even one term in the sequence that is equal to or larger than the previous term, then the sequence is not strictly decreasing.

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