TITLE: Definition of "not converging" and proving $(-1)^n$ does not converge to $1$.
QUESTION [8 upvotes]: Remember that a sequence $x_n, n = 1,2,3\cdots$ is said to converge to $x$ as $n → ∞$ if for all $ε > 0$ there exists an $N ∈ \mathbb{N}$ such that $|x_n − x| < ε$ for all $n ≥ N$.
(a) Complete the following statement:
“If the sequence $x_n, n = 1,2,3\cdots$ does not converge to $x$ as $n → ∞$, that means that there exists an $ε > 0$ such that...”
(b) Consider the sequence $x_n = (−1)^n, n = 1,2,3\cdots$ that is, the sequence is $(−1,1,−1,1,−1,...)$. Prove carefully, starting from your answer to part (a), that this sequence does not converge to 1.
I am confused with the first part and what epsilon represents!

REPLY [3 votes]: Pedro explained clearly the meaning of a sequence not converging here.
Below, I prove that the particular sequence you give doesn't converge to any real number, which implies the particular claim that the sequence doesn't converge to 1.
Claim: $\lbrace (-1)^n \rbrace$ doesn't converge to any number
Tiny Lemma:
$$\forall n \in \mathbb{N} \ (-1)^n \in \lbrace -1, 1\rbrace$$ 
Or, in other words, every term in the sequence equals either $1$ or $-1$.
Proof by contradiction:
Suppose to the contrary that $$\lim_{n\to\infty} (-1)^n = L$$.
Then by the definition of limits, we can choose $$N \in \mathbb{N}$$ such that $$\forall n \in \mathbb{N} \ n \geq N \implies |(-1)^n - L| < 1 $$
In particular, we have that:
  $$|(-1)^N - L| < 1$$
  and
$$|(-1)^{N + 1} - L| < 1$$
Since either $(-1)^N = 1$ or $(-1)^N = -1$ and, either way, $(-1)^{N+1}$ has the opposite sign, we can conclude:
$$ \tag{1a} |1 - L| < 1$$
and
$$ \tag{2a} |-1 - L| < 1$$
By the definition of absolute value, this implies:
$$ \tag{1b} L - 1 < 1 < L + 1$$
and
$$\tag{2b}L - 1 < -1 < L + 1$$
By adding $-1$ to $(1b)$ we can see that 
$$\tag{1c} 0 < L$$
and by adding $1$ to ${2b}$
$$\tag{2c} L < 0$$
which is a contradiction!