How to prove $i^n$ is divergent

Question

How do I prove that $s_n=i^n$ is divergent in Complex space ?

Proving something not divergent is slightly difficult. Because I have to find two parameters $\epsilon$ and $n \geq N$ given $N$ and any $s$

Answer 1

Be careful when you take the negation of the definition of convergence:

For any potential limit $L$, there exists a gap $\epsilon >0$ such that for any $N$, there is some $n>N$ for which $|x_n - L| > \epsilon$.

This approach needs you to show that it does not converge to any limit value.

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Another way: If going with marty's answer, you would want to negate the definition of Cauchy sequence:

There exists $\epsilon$ such that for any $N$, there exists $m,n>N$ for which $|x_n - x_m| > \epsilon$ . In this case you can take $\epsilon$ to be anything $<1$, like $1/2$.

... and if the sequence is not Cauchy, then it cannot converge.

Answer 2

$|i^n-i^{n+1}|>1$, so it can never converge.