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Previous related question.

After I negated the definition of a convergent sequence, I ended up with the following mathematical statement:

$$\exists\ \epsilon > 0,\ \forall\ N \in \mathbb R\ \exists\ \mathbb N \ni n > N : |x_n - l| \ge \epsilon$$

Is this correct? I'd like clarification...

Anyway, I'm now asked to use my negation of the definition of convergence to prove that $(a_n)=((-1)^nn)$ is divergent...

Am I right in assuming it's enough to prove that there exists at least one $\varepsilon$ such that $|(-1)^nn -\mathscr l|\ge\varepsilon$? Can I set $\mathscr l$ to be any real number I please?

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Vladimir Nabokov
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  • "Is this correct?" Hardly. And this does not correspond to what is written on the other page. – Did Oct 30 '13 at 16:33

1 Answers1

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Hints:

  • Convergence to $\ell$ (your previous question): $$ \forall\varepsilon\gt0,\ \exists N\in\mathbb N,\ \forall n\in\mathbb N,\ n\geqslant N\implies |x_n-\ell|\leqslant\varepsilon$$
  • Convergence: $$ \exists\ell\in\mathbb R,\ \forall\varepsilon\gt0,\ \exists N\in\mathbb N,\ \forall n\in\mathbb N,\ n\geqslant N\implies |x_n-\ell|\leqslant\varepsilon$$
  • If convergence then: $$\exists N\in\mathbb N,\ \forall n\in\mathbb N,\ \forall k\in\mathbb N,\ n\geqslant N, k\geqslant N\implies |x_n-x_k|\leqslant1$$
  • If convergence then: $$\exists M\in\mathbb R,\ \forall n\in\mathbb N,\ |x_n|\leqslant M$$

Can you prove that these hold and use one of these to show that $a_n=(-1)^nn$ defines a divergent sequence $(a_n)$?

Did
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