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 $\ds \lim_{n \mathop \to \infty} z_n$ 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
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 4$. Elementary Functions of a Complex Variable: Exercise $1 \ \text {(ii)}$