Convergence of Sequences: Confusion in definition This is the definition of convergence we've been taught -
$(x_{n}) \rightarrow L$ if for every $\epsilon>0 , \exists N \in \mathbb{N}$ such that for all $n \geq N$ it follows that $|x_{n}-L|< \epsilon$
I don't understand the importance of the part in bold, i.e. why is it necessary for all n?
Please provide a counterexample (a case which shows the necessity of the text in bold) if possible! Thank you!
P.S. I feel it should be okay if for certain values of n the terms of the sequence don't lie in the designated interval, most other terms can converge to L. What am I missing?
P.P.S.
I guess my question boils down to the necessity of monotonicity for convergence.
Here's an example to provide more clarity into what I'm thinking:
$(x_{n})= 1/2^n$ for all $n$, except $n=4$. I separately define $x_{4}= 5$.
For very large $n$, this sequence approaches $0$.
 A: The basic issue is with finite vs. infinite # of such exception values. With any finite number, there will always be a maximum value among this set, call it $M$. Thus, for any $N \gt M$, it'll then be true that the inequality of $|x_n - L| \lt \epsilon$ holds for all $n$ $\ge N$.
However, if there's an infinite # of such exception values, then you can't really say that the limit of $x_n$ is $L$ since it doesn't consistently "approach" the value, i.e., there's some $\epsilon \gt 0$ such that, no matter how large your $N$ is, there will always be an infinite number of $n \ge N$ such that $|x_n - L| \ge \epsilon$. This goes counter to the basic concept of what a "limit" is supposed to mean, e.g., it doesn't "converge" as stated in the question comment by coffeemath.
A: Example: $(-1)^n$. Obviously(intution is enough), it isn't convergent. But for any $N$ you can find $n>N$, such that $|(-1)^n-1|=0<\epsilon$. In fact, you can find infinitely many $n>N$ such that $|(-1)^n-1|=0<\epsilon$. So, the for all is really important.
