Definition of "not converging" and proving $(-1)^n$ does not converge to $1$. 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!
 A: This addresses part a):
Logical Preliminaries
A biconditional $P\Leftrightarrow Q$ ($P$ iff $Q$) is equivalent to its contrapositive $\sim Q\Leftrightarrow\sim P$ (not $Q$ iff not $P$).
The negation of a universal $\sim\forall x,P$ (it is not the case that for all $x$, $P$) is $\exists x:\sim P$ (there is an $x$ such that not $P$).
The negation of an existential $\sim\exists x:P$ (it is not the case that there is an $x$ such that $P$) is $\forall x,\sim P$ (for all $x$, not $P$).

Application
The original statement is
$$
\left(\forall\epsilon\gt0,\exists N\in\mathbb{N}:\forall n\ge N,|x_n-x|\lt\epsilon\right)\iff\text{$x_n$ converges to $x$ as $n\to\infty$}
$$
Its contrapositive is
$$
\text{$x_n$ does not converge to $x$ as $n\to\infty$}\iff\left(\exists\epsilon\gt0:\forall N\in\mathbb{N},\exists n\ge N:|x_n-x|\ge\epsilon\right)
$$

Part b) is to apply the forgoing to show non-convergence.
A: 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!
A: Convergence is stated as
"For every $\epsilon >0$ there exists a natural number $N$ such that $n\geq N$ implies $|x-x_n|<\epsilon$"
You might write it as $$\forall\epsilon >0\;\;\exists N\in\Bbb N \;\;\forall n\geq N \text{ we have } |x_n-x|<\epsilon$$
Now, we need to think, when can the above be false? We need just a "counterexample", that is, an $\epsilon >0$ for which no $N$ will every $|x-x_n|<\epsilon$, even though we make $n\geq N$. We might write this as
"There exists an $\epsilon >0$ such that for every natural number $N$, there exists an $n\geq N$ with $|x-x_n|\color{red}{\geq} \epsilon $."
Can you try and prove why $(-1)^n\not\to 1$? Hint: Take $\epsilon =1/2$ in the defintion.

ADD Alternatively, we can think about convergence as follows. Let's define the set 
$$B(x,\epsilon)=\{y\in\Bbb R:|x-y|<\epsilon\}$$
This is usually called "the open ball with center $x$ and radius $\epsilon$. In $\Bbb R$ it is an open interval $(x-\epsilon,x+\epsilon)$, but in $\Bbb R^2$ it is a disk (with the Euclidean metric) and in $\Bbb R^3$ is a ball (a filled sphere). Now, we may state convergence as follows.
DEF Let $\langle x_n:n\in\Bbb N\rangle$ be a sequence in $\Bbb R$. Let $x\in \Bbb R$. We say that $x_n$ converges to $x$ if for each ball $B(x;\epsilon)$ we're given, there exists an $N$ such that the tail sequence 
$$\langle x_n:n\geq N\rangle=\langle x_N,x_{N+1},\dots\rangle$$
is contained entirely in $B(x,\epsilon)$. 
This definition helps in the sense that we can see convergence fails when we can find some $\epsilon>0$ such that no matter which "tail" ($N$ big) we take, some element of it will fail to be inside the ball $B(x;\epsilon)$. This directly generalizes to $\Bbb R^n$ with $$\|{\bf x}-{\bf y}\|:=\left(\sum_{i=1}^n (x_i-y_i)^2\right)^{1/2}$$
and $$B({\bf x};\epsilon):=\{{\bf y}\in\Bbb R^n:\|{\bf x}-{\bf y}\|<\epsilon\}$$
A: Here is (a).
(i.o means "infinitely often")
$$\sim \exists x \forall \epsilon > 0 \exists N {\rm s.t.} n\ge N \Rightarrow d(x_n, x) < \epsilon \iff \forall x \exists \epsilon > 0\; {\rm s.t.}\;
d(x_n, x) \ge \epsilon\; {\rm i.o}$$
More digestibly, for any real $x$ there is an open interval $I$ containing $x$ so that $x_n\not\in I$ infinitely often.
