Is a convergent sequence eventually constant?

Is a convergent sequence eventually constant?

Any eventually-constant sequence is convergent (hence Cauchy). A sequence (xn)n∈N is eventually constant iff there is some x and some N so that n ≥ N guarantees xn = x.

What is an eventually constant sequence?

(a) A sequence {xn} in a metric space is called eventually constant if there exists some N such that for all n>N, xn = p for some p ∈ M. Show that any eventually constant sequence converges.

Is every convergent sequence bounded?

Theorem 2.4: Every convergent sequence is a bounded sequence, that is the set {xn : n ∈ N} is bounded. Remark : The condition given in the previous result is necessary but not sufficient. For example, the sequence ((−1)n) is a bounded sequence but it does not converge.

Is every decreasing sequence convergent?

Informally, the theorems state that if a sequence is increasing and bounded above by a supremum, then the sequence will converge to the supremum; in the same way, if a sequence is decreasing and is bounded below by an infimum, it will converge to the infimum.

Do sequences converge?

A sequence is said to be convergent if it approaches some limit (D’Angelo and West 2000, p. 259). Every bounded monotonic sequence converges. Every unbounded sequence diverges.

How do you prove a sequence is decreasing?

If anIf an>an+1 a n > a n + 1 for all n, then the sequence is decreasing or strictly decreasing .

What is a decreasing sequence?

From Encyclopedia of Mathematics. A sequence {xn} such that for each n=1,2,…, one has xn>xn+1. Sometimes such a sequence is called strictly decreasing, while the term “decreasing sequence” is applied to a sequence satisfying for all n the condition xn≥xn+1.

What makes a sequence converges?

A sequence is “converging” if its terms approach a specific value as we progress through them to infinity.

How do you prove that (x n) is convergent?

If ( x n) is bounded, it is necessarily the case that ( x n), given that it is monotone increasing, IS convergent (you can prove this). So, ∃ N ∈ N such that ∀ n ≥ N it follows that x n > b. Given that ( x n i) is bounded above by b, this means that ∀ i ∈ N, n i < N.

How do you prove that a sequence converges to a?

Then for every x after y k in original sequence it’s also true since sequence is monotone.Also, for every x there exists an y which stands further in ( x n). Then every x is < A. Then the sequence converges to A. (It was for increasing sequence, for decreasing anlogically).

How to find the greatest lower bound of a decreasing sequence?

For decreasing sequences we have the following result and its proof is similar. Theorem 2.6: Suppose (x n) is a bounded and decreasing sequence. Then the greatest lower bound of the set fx n: n2Ngis the limit of (x n). Examples: 1. Let x 1 = p 2 and x n = p 2 + x n 1 for n>1:Then use induction to see that 0 x n 2 and (x

Is (x n) a monotone increasing sequence?

Given its monotonicity, it follows that ( x n) is unbounded, that is, ∀ M ∈ N, ∃ N ∈ N such that ∀ n ≥ N it follows that x n > M (at some point ” N ,” the sequence passes the boundary M for any boundary M ∈ N). If ( x n) is bounded, it is necessarily the case that ( x n), given that it is monotone increasing, IS convergent (you can prove this).