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

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The sequence defined by \( a_n = 2 - (-1)^n \) does not converge, as established through rigorous definitions of convergence. The sequence oscillates between the values 1 (for even n) and 3 (for odd n), demonstrating that for any proposed limit \( \ell \), there exist infinitely many terms such that \( |a_n - \ell| > 1 \). This confirms that the sequence fails to meet the criteria for convergence and is not a Cauchy sequence.

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I have that the sequence a_n=\{2-(-1)^n\} not converges. I must show this with the rigorous definition.

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

How i can continue?
 
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Try showing that it's not a cauchy sequence instead and then just say "therefore it is not convergent."
 
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Alternatively, note that a_n= 1 for n even, a_n= 3 for n odd. For any l, there exist arbitarily large n such that |a_n- l|> 1, half the distance between 1 and 3.
 


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

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


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.
 

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