4
$\begingroup$

Let $$ (a_n)_{n \in \mathbb{N}}$$ be a sequence.

According to the definition of limit, it is said that $$ \lim_{n \rightarrow \infty} a_n=L \Leftrightarrow \left(\forall \varepsilon>0, \ \exists n_0 \in \mathbb{N}, \ n \ge n_0 \Rightarrow \left|a_n-L \right|<\varepsilon \right). $$

About the details, does it really matter, whether you say $$ n \ge n_0 \ \mathrm{or} \ n > n_0 ?$$

And why would $$ n_0 $$ have to be an integer?

$\endgroup$

2 Answers 2

4
$\begingroup$

We don't need that $n_0$ is an integer and it doesn't matter whether we use $\ge$ or $>$ both lead to an equivalent definition.

$\endgroup$
3
$\begingroup$

For the first part of your question, no it does not particularly matter as both statements will hold true for some epsilon greater than 0.

$n_o$ refers to the $n_o$ term of the sequence and thus is an integer. The limit $L$ however can be a real number.

$\endgroup$
1
  • 1
    $\begingroup$ Not necessarly $n_0$ needs to be an integer, the definition works fine alfo with $n_0 \in \mathbb R$. But of course we can also use $n_0 \in \mathbb N$ as in te given example. $\endgroup$
    – user
    Sep 8, 2020 at 13:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .