Question

can anyone help me to use subsequences to prove the convergence of a sequence?

Answer 1

What type of sequences are we talking about...monotone?

Answer 2

Is this a proof question or a specific-sequence|determine-convergence question?

Answer 3

it is a proof question. there is an arbitrary sequence \[x _{n}\] that has exactly one cluster point. i need to prove that it is convergent

Answer 4

Alright, I need the prompt....Exactly as they give it.

Answer 5

"suppose that x(n) (imagine that's a subscript n) is a bounded sequence of distinct real numbers such that its range \[x _{n} : n \epsilon \mathbb{N}\]
has exactly one cluster point. prove that x(n) is convergent"

Answer 6

Given:
\[(1)\;\;\;\;x_{n} \;\;is\;\; bounded.\]
\[(2) x_{n}\;\;has\;\;exactly\;\;one\;\;cluster\;\;\; point.\]
Proof:
\[By \;\;the\;\; completeness \;\;axiom \;\;\;theorem, \;\;x_{n} \;\; is \;\;bounded\;\;above \;\; and\;\; below.\] So there is a point M and P s.t \[M\le x_{n} \;\;\;and\;\;\;\;\; P \ge x_{n}\;\;\;for\;\;all \;\;\;\;n\in\aleph \]

Note: I do not know how you have defined cluster points in your class, but I will take a guess.

By definition of a cluster point, for any \[\epsilon >0\] there exist an \[m \in \aleph\;\;\; s.t.\;\;\;m>n\] and for all m

\[|{x_{n}-x_{m}}|<\epsilon\]

Formatting latex in this website is killing me, I hope this helps you somewhat....I need to talk to the owners of this website.

Answer 7

true story, the formatting is balls to the wall stupid. but thank you so much!! i shall now spend the next few hours studying this in my math shrine