epsilon-delta proofs 4

In [Def-1] we defined a real number sequence $a_n$ approaching $a$. Although [Def-1] is much complicated, the reason why we should use it is that we were able to discuss precisely.

Using mathematical symbols, [Def-1] is also expressed as follows.

$a_n$ approaches $a$, when
$$(\forall \epsilon>0)(\exists N\in \mathbb{N})(\forall n\in \mathbb{N})(n\geq N\rightarrow |a_n-a|< \epsilon)$$
where $\forall$ means 'all ' or 'any', $\exists$ means 'exist ' or 'some', and $\in$ means 'in elements of '.

Make a contrapositive statement of [Def-1] forcibly.  Next sentence is the answer. If
  $$(\exists \epsilon>0)(\forall N\in \mathbb{N})(\exists n\in \mathbb{N}) (n\geq N \land |a_n-a|\geq \epsilon),$$
a real number sequence $a_n$ never approaches $a$, where $\land$ means 'and '.

That is, $a_n$ does not approach $a$ when for some $\epsilon>0$, any $N>0$, there is $n>0$ such that $n\geq N$ and $|a_n-a|\geq \epsilon$.

The sequence $2.3888\cdots$ does not approach $2.4$. But the sequence $2.3888\cdots$ approaches $43/18$. One of the sequence which approaches $2.4$ is $2.3999\cdots$. These shows that the converging depends on the relation between a sequence $a_n$ and a value $a$.

We are interested in the case in which $a_n$ does not converge with any values. It is called diverging. In particular, if the greater $n$ becomes, the greater $a_n$ becomes, we say that $a_n$ approaches infinity, or the limit of $a_n$ is $\infty$ and expresses that
$$\lim_{n\rightarrow \infty} a_n =\infty, \quad or \quad a_n\to\infty \quad (n\to \infty )$$
Next definition is equivalent to above.

[Def-3] $a_n$ approaches infinity, when for any $\epsilon >0$, there is $N>0$ such that if $n\geq N$, $|a_n-a|\geq \epsilon$.