all but finitely many terms

Question

So I found a definition in my notes for a sequence to be convergent. It said that a sequence is said to converge to a real number $L$ if every open interval containing $L$ contains all but finitely many terms of this sequence $(a_n)_{n\in\Bbb N}$. What does it mean to have all but finitely many terms? I thinks there may be just finitely many terms??

Answer 1

Given an open interval $U$ containing $L$, there exists $N$ such that $n\geq N$ implies that $a_n\in U$.

Answer 2

If your sequence is $(x_n)_{n\in\mathbb N}$, asserting that a set $O$ contains all but finitely many terms of the sequence means that the set $\{n\in\mathbb{N}\,|\,x_n\notin O\}$ is finite.

Answer 3

All but finitely many terms means all $a_n$ such that $n$ is an integers greater or equal to $k$ for some positive integer $k$.

Since $k$ is finite all but finitely many integers are greater than or equal to $k$.

Answer 4

As an example, imagine an infinite stream of $1$s and $0$s, basically a sequence: $1, 0, 1,1, \dots $

If there are now an infinite amount of $1s$, but finitely many $0$s, then you could say all, but finitely many elements are equal $1$. This means: You can walk along this sequence, you would encounter $1$s and $0$s.

However after some point, you will meet the last $0$ (this is a specific element in your list!) and thus all elements following would be $1$.

We can now translate this to your example: You have a sequence $a_n$, that may converge to $L$. Now you pick any open interval $I$ containing $L$. We can then come back to our sequence and go through it and ask for every element: Is it inside $I$?

If yes, then we think of it as a $1$, if no, we think of it as a $0$.

Your definition now tells us: The sequence will only converge to $L$ if at one point the $0$s will stop and we encounter $1$s for any open interval $I$ - how abitrarily small we choose it to be.