Your definition of convergence is not correct. It should be

**There exists** L in Real numbers **such that for** all E>0 such that there exists N in Natural numbers **such that** n>/= N implies |xn-L|**<** E.

The negation of the definition of convergence should be: For every real number $\displaystyle L$ there exists a real number $\displaystyle \varepsilon>0$ such that for every natural number $\displaystyle N,$ there is a natural number $\displaystyle n\geqslant N$ such that $\displaystyle \left|x_n-L\right|\geqslant\varepsilon.$ Use this to prove the divergence of $\displaystyle \left(x_n\right)_{n=1}^\infty$ where $\displaystyle x_n=(-1)^n.$ Hint: Given $\displaystyle L,$ let $\displaystyle \varepsilon=\min\left\{|L+1|,|L-1|\right\}$ if $\displaystyle L\ne\pm1$ and $\displaystyle \varepsilon=2$ if $\displaystyle L=\pm1.$