Going Beyond

To find the sum of $1-1+1-1+...$ we take the artithmetic mean of the cycling results, by which we obtain $1/2$ . Using this logic, what is the value we obtain when evaluating $i$ taken to successive powers of integers towards infinity?

Moderator's edit: I think he/she wants to calculate the Cesaro sum of $\displaystyle \lim_{n\to\infty} \dfrac1n\sum_{k=1}^n i^k$ where $i = \sqrt{-1}$ .