Proving the sequence $(-1)^n$ is divergent by the formal definition I know that the definition of a convergence sequence definition is:
$$(\exists L\in \mathbb{R})(\forall\varepsilon > 0)(\exists N \in \mathbb{N})(\forall n\in\mathbb{N})[n \ge N \Rightarrow |x_n-L| < \varepsilon]$$
And by using the negation, we can obtain the divergence sequence definition:
($\forall$ L $\in$ $\mathbb{R}$)($\exists$ $\epsilon$ > 0)($\forall$ N $\in$ $\mathbb{N}$)($\exists$ n $\in$ $\mathbb{N}$)[n $\ge$ N and |$x_n-L|$ $\ge$ $\epsilon$]
So, as far as I understand, for any real number "L" I can find an epsilon such that for any natural number, the following holds true: |$x_n-L|$ < $\epsilon$
Since we have $a_n=(-1)^{n}$, the sequence alternates between 1 and -1.
If we say L = 1 (From the definition the inequality is valid $\forall$ L $\in$ $\mathbb{R}$).
Then we have: |$x_n-1|$ $\ge$ $\epsilon$, but as the sequence is 1 or -1, if we plug 1:
|1-1| $\ge$ $\epsilon$ => $0$ $\ge$ $\epsilon$ > $0$, which is not valid because from the definition $\epsilon$ > $0$. What am I doing or understanding wrong?
 A: When verifying a quantified definition like that of a divergent sequence, you should treat variables following "for all" as being given to you - you have no say in how they are chosen. Variables following "there exists" may be chosen by you using any previously established variables.
Read the definition of divergent sequence left-to-right:

*

*for every $L \in \mathbf R$:  a value of $L$ is given to you. You don't know anything else about it.

*there exists $\epsilon > 0$: we get to pick this one. How about $\epsilon = 1$.

*for every $N \in \mathbf N$: again this is given to you. You don't get to define it.

*there exists $n \in \mathbf N$: we get to pick this one too. Its value can depend on $L$, $\epsilon$, and $N$ if necessary. How about $n = 2N$ if $L < 0$ and $n = 2N+1$ if $L \ge 0$.

Then:

*

*$|(-1)^n - L| = |1-L| > 1$ if $L < 0$, and

*$|(-1)^n - L| = |(-1) - L| \ge 1$ if $L \ge 0$.

In both cases you have $n \ge N$ and $|(-1)^n - L| \ge \epsilon$. This verifies the definition.
A: As you mentioned in your definition of divergence, you only need to prove that there exists SOME $n$ larger than $N$ for which $\mid a_n - L \mid > \epsilon$ where $\epsilon$ is a positive number of choice. To prove this sequence diverges, we can select $\epsilon = 0.5$.
If you want to prove the series does not converge to $L=1$, you only need to select an odd $n$ larger than $N$. Then, $a_n = -1$ and its difference from $L$ exceeds $\epsilon$(This is of course not true for ALL $n\geq N$).
To prove the series does not converge for any choice of $L$, again you can choose $\epsilon = 0.5$. This would imply that for any $N$, all $a_n$'s where $n \geq N$ have the property that $\mid a_n - L \mid < 0.5$. But that would imply:
$$ \mid 1 - L \mid < 0.5 $$
$$ \mid -1 - L \mid < 0.5 $$
This is impossible as no real number satisfies the two said inequalities simultaneously. Therefore, this sequence does not converge.
