TITLE: Divergence of a sequence.
QUESTION [5 upvotes]: In an example in the book, Thomas Calculus 14e:
Q: Show that sequence $\{(-1)^{n+1}\}$ diverges?
A: They choose $\varepsilon$ to be $1/2$ and thus,
$|L-1|<1/2$ for $+1$ and $|L+1|<1/2$ if value converges to $+1$, where $L$ is the value to which the sequence converges. On solving both equations, we get no common solution, thus no such limit $L$ exists and thus the sequence diverges.
My question: If we choose $\varepsilon$ to be $3$ or any number greater than it, then the above inequalities give a solution, so does it mean the sequence converges??
Also later it states, a sequence can diverge without diverging to $\pm \infty$, how is that relevant and possible?

REPLY [3 votes]: A sequence  of real numbers $\{a_n\}$ converges to some $a\in\mathbb R$ if given any $\varepsilon\gt 0$, there exists some  $N\in\mathbb N $ such that  $|a_n-a|\lt \varepsilon$ for all $n\geq N $. 
Suppose the sequence  $\{(-1)^{n+1}\}$ converges to some  $L\in\mathbb R$. So,in particular, given  $\varepsilon=\frac 12$, we should be able to find some $N$ such that $|L-(-1)^{n+1}|\lt\frac 12$ for all $n\geq N $. That is, $|L+1|\lt \frac 12$ and $|L-1|\lt \frac 12$. In other words, $\frac {-3}2\lt L\lt \frac {-1}2$ and  $\frac 12\lt L\lt \frac {3}2$, which is not possible. 
You can do a similar kind of argument with any $\varepsilon$ such that  $0\lt \varepsilon \lt 1$. And notice that the sequence does not go to infinity, but oscillates between  $1$ and $-1$.