Divergent Complex Sequence/Examples/i^n

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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.

$\blacksquare$


Sources