Divergence of a sequence.

No.

The definition of convergence states that it is true for any $\epsilon > 0$, not just for a particular one.

So, if the definition of convergence fails for a particular $\epsilon$, then the sequence diverges.

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 $-1$.