Question:Prove that this sequence does not converge for any $x \in \Bbb R.$


Definition (Negation):$ \exists \varepsilon \forall N \in \Bbb N \exists n \in \Bbb N, n>N: |x_n-x|\geq \varepsilon$

I would usually tackle this question by taking the odd/even sub-sequence and showing that the limits are unequal, however I am struggling with proving it using the negation for the definition of convergence. Mainly, what is the logic of picking epsilon/n?

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    $\begingroup$ What are the limits of those subsequences, and what is the distance between those limits? Take $\epsilon$ any number larger than $0$ and smaller than that distance. $\endgroup$ – logarithm May 12 at 14:17
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    $\begingroup$ In this question, unlike the previous one, you are not disproving that some specific number is the limit, but proving that no number is the limit. In this case it is useful to note that $|x_n-x|<\epsilon$ and $|x_{n+1}-x|<\epsilon$ would imply $|x_n-x_{n+1}|\leq |x_n-x|+|x-x_{n+1}|<2\epsilon$. So, you would try to disprove $|x_n-x_{n+1}|<2\epsilon$, which is an inequality that doesn't mention any specific candidate $x$ for the limit. Here we used $x_n$ and $x_{n+1}$ because you are going to look at even and odd terms. $\endgroup$ – logarithm May 12 at 14:25
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    $\begingroup$ In general you would try to disprove that the sequence is a Cauchy sequence. The Cauchy property is useful because it doens't mention any particular limit, and being equivalent to being convergent for real numbers, allows you to complete the proof. If the sequence is not Cauchy, then it can't be convergent. $\endgroup$ – logarithm May 12 at 14:27
  • $\begingroup$ Thank you that makes sense. $\endgroup$ – ViB May 12 at 14:34
  • $\begingroup$ Hi @logarithm just spent quiet a while trying to figure this out. After subbing in $x_n$ and $x_{n+1}$ I got $|2-\frac{1}{n}-\frac{1}{n+1}|$. If I prove that $|2-\frac{1}{n}-\frac{1}{n+1}| \geq 2\varepsilon$ then this would essentially be disproving the inequality in you comment right? $\endgroup$ – ViB May 12 at 16:07

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