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My task is to write a precise mathematical statement that "the sequence $(a_n)$ does not converge to a number $\mathscr l$"

So, I have my definition of a convergent sequence: "$\forall\varepsilon>0$ $\exists N\in\Bbb R$ such that $|x_n -\mathscr l|<\varepsilon$ $\forall n \in \Bbb N$ with $n>N$"

Would the correct negation of this be "$\forall\varepsilon>0$ $\exists N\in\Bbb R$ such that $|x_n -\mathscr l|>\varepsilon$ $\forall n \in \Bbb N$ with $n>N$"?

It doesn't seem that this is the answer as the next part of my task is to prove that a sequence is divergent using my formed proof, but it'd be difficult to do since it's a general proof of divergence and not just a proof that $(a_n)$ doesn't converge a specific number $\mathscr l$

Perhaps I should find a prove that $(a_n)$ tends to $\pm\infty$? This is more simple but it does not include monotone sequences such as $x_n:=(-1)^n$.

Can someone assist me with this task? All comments and answers are appreciated.

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    $\begingroup$ A sequence $(a_n)_{n\in \Bbb N}$ converges to $1$ if, and only if, $$\forall \varepsilon \exists N\in \Bbb N\Bbb \forall n\in \Bbb N(n\ge N\implies |a_n-1|<\varepsilon).$$Apparently you're using $N\in \Bbb R$ andusing the strict inequality in the antecedent of the implication, that's fine, Do you know how to negate something that looks like $\forall yP(y)$ or $\exists yP(Y)$? $\endgroup$ – Git Gud Oct 30 '13 at 15:31
  • $\begingroup$ Yes I do. Shall I use that? $\endgroup$ – Vladimir Nabokov Oct 30 '13 at 15:44
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    $\begingroup$ Yes. Read Pete's answer for further details. $\endgroup$ – Git Gud Oct 30 '13 at 15:48
  • $\begingroup$ How about this as the negation of the definition of a convergent sequence: "$\exists \varepsilon > o \forall N\in\Bbb N$ such that $|x_n - \mathscr l|>\varepsilon$ with $n>N$" $\endgroup$ – Vladimir Nabokov Oct 30 '13 at 15:52
  • $\begingroup$ 'with' is very unclear. $\endgroup$ – Git Gud Oct 30 '13 at 16:22
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This is not the correct negation. Consider $x_n = (-1)^n$ and $l = 1$. The correct negation can be expressed as $$\exists\ \epsilon > 0,\ \forall\ N \in \mathbb R\ \exists\ \mathbb N \ni n > N : |x_n - l| \ge \epsilon$$

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  • $\begingroup$ @GitGud edited. $\endgroup$ – AlexR Oct 30 '13 at 15:40
  • $\begingroup$ The inequality shouldn't be strict and if you wanna use the same definition as the OP, $N$ should be quantified over $\Bbb R$. It also wouldn't hurt to specify that $n$ is a natural number. $\endgroup$ – Git Gud Oct 30 '13 at 15:42
  • $\begingroup$ Just for the sake of completeness, the statements are (trivially) equivalent, but I have nonetheless adapted your suggestions. $\endgroup$ – AlexR Oct 30 '13 at 16:58
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    $\begingroup$ I agree that they are equivalent, but these details make the difference to someone who's learning, that's why I pointed them out. (+1) $\endgroup$ – Git Gud Oct 30 '13 at 17:20
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No, what you've written is not correct.

It looks like you need practice negating multiply-quantified statements. The key idea is that when you move a negation past a quantifier, it flips the quantifier from universal to existential or vice versa. Thus for instance

$\neg (\forall x$ $P(x))$

is logically equivalent to

$\exists x (\neg P(x))$.

Here $\neg$ means "not".

It would also be a good exercise to find an example to show that your proposed negation of "$a_n \rightarrow l$" need not be correct. As a hint: your condition implies that the sequence is unbounded.

I didn't really understand the second part of your question. In particular I don't follow "the next part of my task is to prove that a sequence is divergent using my formed proof". Maybe it's best to focus on one question at a time. Once you understand the negation question properly, you can ask the next part as a new question if you like.

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The statement would be something like:

$ \forall L, \exists \epsilon > 0 : \forall N, \exists n>N : |a(n)-L| \geq \epsilon$

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I am just going to define it in words and then explain the formulation.

The definition of convegence say that for every value of $\epsilon > 0$, you will definitely get a natural number $N(\epsilon)$ such that $|x_n-l|<\epsilon$ for all $n \geq N(\epsilon)$.(where $l$ is the point of convergence).

Now, to negate this statement, you just have to find at least one $\epsilon >0 $ for which there exist no $N(\epsilon)$ s.t. $|x_n-l|<\epsilon$ for all $n \geq N(\epsilon)$ holds true. This is equivalent to saying that for all natural numbers $N$ you will get atleast one natural number $M$ s.t. $|x_M-l|>\epsilon$ where $M > N $.

For e.g. You will start with $N=1$ and you will definitely get a natural number $M_1$ greater than 1 for which $|x_{M_1}-l|>\epsilon$. If you start with $N=2$ then you will definitely get a natural number $M_2$ greater than 2 for which $|x_{M_2}-l|>\epsilon$ and this will keep on going for all natural numbers $N$.

The logical statement of this, then, converts to $$ \exists\ \epsilon > 0,\ \forall\ N \in \mathbb N\ \exists\ \mathbb N \ni n > N : |x_n - l| \ge \epsilon $$

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