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I am having trouble understanding this question. What does it mean to negate the limit. Does it mean that n does not have a limit? Also I have done the symbolic negation, is this correct?

$\begin{equation}\exists \epsilon > 0, \forall N \in \mathbb Z, \exists n\in \mathbb Z, n > N \wedge \neg L - \epsilon < a_n < L + \epsilon.\end{equation}$.

Here is the question:

Recall from Calculus the definition of the limit of a sequence $a_n$. We say that the limit of the sequence $a_n$ as $n$ goes to infinity equals $L$ and write:

$lim_{n\rightarrow \infty} a_n = L$

We can write this using quantifiers as follows:

\begin{equation}\label{limit}\forall \epsilon > 0, \exists N \in \mathbb Z, \forall n\in \mathbb Z, n > N \rightarrow L - \epsilon < a_n < L + \epsilon.\end{equation}

Explain in words what the negation of this definition means. Now write the negation of the limit from the given formula.

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  • $\begingroup$ The above formula means: "the sequence $(a_n)$ has limit $L$". $\endgroup$ – Mauro ALLEGRANZA Oct 2 '17 at 14:48
  • $\begingroup$ Its negation raeds: "$L$ is not the limit of $(a_n)$". $\endgroup$ – Mauro ALLEGRANZA Oct 2 '17 at 14:48
  • $\begingroup$ @MauroALLEGRANZA is my negation of the formula viable? $\endgroup$ – Safder Aree Oct 2 '17 at 15:01
  • $\begingroup$ @SafderAree No! The last part is wrong $\endgroup$ – Fakemistake Oct 2 '17 at 15:03
  • $\begingroup$ NO; the def of limit has: $\ldots \forall n > N \ |a_n - L| < \epsilon$. When we negate it we get: $\ldots \exists n > N \ |a_n - L| \ge \epsilon$. The negation of "less than" is "greater-or-equal to". $\endgroup$ – Mauro ALLEGRANZA Oct 2 '17 at 15:10
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The sequence $a_n$ converges (to the limit $L$) if and only if the following is true: $$\exists L\in\mathbb{R}\,\forall \epsilon>0\,\exists N\in\mathbb{N}\,\forall n\geq N\Rightarrow |a_n-L|<\epsilon$$

The negation of $L$ is the limit of the real sequence $(a_n)_{n\in\mathbb{N}}$ is $$\forall L\in\mathbb{R}\,\exists \epsilon>0\,\forall N\in\mathbb{N}\,\exists n\geq N \Rightarrow |a_n-L|\geq \epsilon $$

What I have done so far? I turned $\exists$ into $\forall$ and backwards. Note that $<$ is replaced by $\geq$

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  • $\begingroup$ Would you be able to show me the steps of how you got from the given form to this? $\endgroup$ – Safder Aree Oct 2 '17 at 14:47

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