$2\mathrm{.}2\mathrm{.}2.$ Consider the sequence of real numbers $(z_n)$ defined for each index ninN by

$z_n=\frac{4(-1{)}^n}{\sqrt[\phantom{2}]{n1}}2(-1{)}^n$

Exercise $2\mathrm{.}1\mathrm{.}2$ shows us $2$acl$(z_n)$ and more. Here, prove $\underset{n}{lim}{z}_{n}2.$