Monotonic 0<an<1 for all n and no two terms are the same

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Homework Help Overview

The discussion revolves around finding a sequence {an} that is monotonic, satisfies the condition 0 < an < 1 for all n, has no repeated terms, and converges to 1/2 as n approaches infinity.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants have proposed various sequences, such as 1/(2√n), n/(2n-1), and 1/2^n, but express uncertainty about their validity. One participant suggests a sequence defined as a_n = 1/2 + 1/n for n > 2 and questions its applicability. Others inquire about the initial terms of the sequence and the definition of monotonicity.

Discussion Status

The discussion is active, with participants exploring different sequence formulations and clarifying the concept of monotonicity. There is no explicit consensus on a valid sequence yet, but multiple ideas are being examined.

Contextual Notes

Participants are working under the constraints of the problem statement, including the requirement for the sequence to be monotonic and to remain within the bounds of 0 and 1. The definition of monotonicity is also being discussed to clarify its implications for the sequence.

gregy6196
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Homework Statement



Give an example of a sequence {an}, satisfying the following:
{an} is monotonic
0<an<1 for all n and no two terms are the same
lim(n→∞) an = 1/2

Homework Equations


what is monotonic


The Attempt at a Solution


1/(2√n)
n/(2n-1)
1/2^n

just been trying genearal sequences but none of them work
 
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Does this one work?

[tex]a_n = \frac{1}{2 } + \frac{1}{n}\ \forall n>2[/tex]
 


what would be the first term?
 


Oh, c'mon! Use any two numbers less than 1/2 for a1 and a2.
 


monotonic refers to how it increases. monotonic increasing means each term is greater than or equal to the term before it. monotonic decreasing means each term is less than or equal to the term before it. if it just says monotonic, either situation will work.
 


dirk_mec1 said:
Does this one work?

[tex]a_n = \frac{1}{2 } + \frac{1}{n}\ \forall n>2[/tex]

For all n
 

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