I have that the sequence [itex]a_n=\{2-(-1)^n\}[/itex] not converges.

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In summary, the conversation discusses the convergence of the sequence a_n=\{2-(-1)^n\} and how to prove that it does not converge using the rigorous definition. One suggestion is to show that it is not a cauchy sequence, while another suggestion is to consider the values of the sequence for even and odd terms. HallsofIvy suggests using epsilon as half the value of the difference between the two possible values of a term in the sequence.
  • #1
solakis
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I have that the sequence [itex]a_n=\{2-(-1)^n\}[/itex] not converges. I must show this with the rigorous definition.

I think use [itex]\exists{\epsilon>0}\forall{N\in\mathbb{N}}\exists{n\geq N}:|a_n-\ell|\geq\epsilon[/itex]

How i can continue?
 
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  • #2


Try showing that it's not a cauchy sequence instead and then just say "therefore it is not convergent."
 
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  • #3


Alternatively, note that [itex]a_n= 1[/itex] for n even, [itex]a_n= 3[/itex] for n odd. For any l, there exist arbitarily large n such that [itex]|a_n- l|> 1[/itex], half the distance between 1 and 3.
 
  • #4


HallsofIvy said:
Alternatively, note that [itex]a_n= 1[/itex] for n even, [itex]a_n= 3[/itex] for n odd. For any l, there exist arbitarily large n such that [itex]|a_n- l|> 1[/itex], half the distance between 1 and 3.

So which do you think should be the value of ε>ο ??
 
  • #5


solakis said:
So which do you think should be the value of ε>ο ??

HallsofIvy just told you what epsilon to use, half the value of the difference of the two possible values any term of the sequence can have.
 

What is the sequence [itex]a_n=\{2-(-1)^n\}[/itex]?

The sequence [itex]a_n=\{2-(-1)^n\}[/itex] is a mathematical sequence in which each term is defined as 2 minus the alternating value of -1 raised to the nth power. This means that the sequence alternates between the values 3 and 1.

Why is it important to determine if the sequence converges or not?

Determining if a sequence converges is important because it helps us understand the behavior of the sequence and make predictions about its future values. Convergent sequences have a finite limit, while non-convergent sequences do not have a limit and may exhibit unpredictable behavior.

How can we determine if [itex]a_n=\{2-(-1)^n\}[/itex] converges or not?

We can determine the convergence of a sequence by taking the limit as n approaches infinity. If the limit exists and is a finite value, then the sequence converges. If the limit does not exist, then the sequence does not converge.

What is the limit of the sequence [itex]a_n=\{2-(-1)^n\}[/itex] as n approaches infinity?

The limit of the sequence [itex]a_n=\{2-(-1)^n\}[/itex] as n approaches infinity does not exist. This can be seen by taking the limit of the even and odd terms separately. The limit of the even terms is 3, while the limit of the odd terms is 1. Since these limits do not converge to the same value, the overall limit of the sequence does not exist.

Why does the divergence of [itex]a_n=\{2-(-1)^n\}[/itex] make sense intuitively?

Intuitively, the sequence [itex]a_n=\{2-(-1)^n\}[/itex] does not converge because it oscillates between two values, 3 and 1, as n increases. This means that the terms of the sequence do not approach a specific value and instead continue to alternate between two values. Therefore, the sequence does not have a limit and does not converge.

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