# Thread: Need help understanding "limit of a sequence" definition

1. ## Need help understanding "limit of a sequence" definition

The picture of the definition is enclosed in the attachments:

I don't understand what the whole greek letter epsilon and this introduced letter "M" mean.

Please reply as if you're talking to a kindergartner because I really don't understand this complicated mathematical jargon.

Thanks!

2. Originally Posted by RedSwiss
The picture of the definition is enclosed in the attachments:

I don't understand what the whole greek letter epsilon and this introduced letter "M" mean.

Please reply as if you're talking to a kindergartner because I really don't understand this complicated mathematical jargon.

Thanks!
that means that every number from sequence $a_n$ except of them finite many, are in $\varepsilon$ region of the number L. actually definition of the sequence...

$\displaystyle (\forall \varepsilon >0) ( \exists n_0 \in \mathbb{N}) (\forall n \in \mathbb{N}) (n\ge n_0 \Rightarrow |x_n-a| < \varepsilon )$

this up there is same as if you write

$\displaystyle \lim_{n\to \infty} x_n = a$

if there is $a\in \mathbb{R}$, so that $\displaystyle \lim_{n\to \infty} x_n = a$ then we say that sequence $x_n$ converge.

if you look at sequence $(-1) ^n$ you know that it's not converge (it diverges) but let's try to show that

okay, let's assume that converges, and that there is for some $a\in \mathbb{R}$ we have that $\displaystyle \lim_{n\to \infty} x_n = a$. if we look at members of the sequence we see that all of them can be (1) or (-1), that means that both of those numbers have to be in any $\varepsilon$ region of the point a. but that is not possible if you chose that your $\varepsilon < \frac {1}{2}$ there is no way to both of those numbers to be in that interval $(a-\varepsilon , a+ \varepsilon )$ which length is less the one

"we say that sequence $x_n$ converges to the point $a \in \mathbb{R}$ if in every $\varepsilon$ region of the point $a$ are almost all members of the sequence "