Going Beyond

Calculus Level pending

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 \(i = \sqrt{-1} \).

×

Problem Loading...

Note Loading...

Set Loading...