Divergent Complex Sequence/Examples/i^n

Example of Divergent Complex Sequence
Let $\sequence {z_n}$ be the complex sequence defined as:
 * $z_n = i^n$

Then $\displaystyle \lim_{n \mathop \to \infty} z_n$ does not exist.

Proof
Consider the subsequence $\sequence {x_{n_r} }$ of $\sequence {x_n}$ where $\sequence {n_r}$ be the sequence in $\N$ defined such that $n_r = 2 r$.

Then $\sequence {x_{n_r} }$ is the real sequence defined as:
 * $x_{n_r} = \paren {-1}^r$

As can be seen in Divergent Sequence may be Bounded, $\sequence {\paren {-1}^r}$ does not converge to a limit.