# Sum of powers of i

What is $1 + i + i^2 + i^3 + i^4 + i^5 + \cdots$?

Of course it’s $1/(1-i)$, right, by the usual formula for summing a geometric series? But this says that

$1 + z + z^2 + \cdots = {1 \over 1-z}$

when $|z|<1$.  And $|i| = 1$, so it doesn’t work here. But who cares? Start taking partial sums. The sum is (after simplifying using $i^2 = -1, i^4 = 1$):

$1 + i - 1 - i + 1 + i - 1 - i + \cdots$

and we can write down partial sums: $1, 1+i, i, 0, 1, 1+i, i, 0, \cdots$ — and the average of this series is $(1+i)/2$, which is $1/(1-i)$. It’s a complex version of Grandi’s series ($1 + 1 - 1 + 1 - 1 + 1 \cdots = 1/2)$, and indeed the argument I’ve outlined here is Cesaro summation.)

## 2 thoughts on “Sum of powers of i”

1. […] Sum of powers of i (gottwurfelt.wordpress.com) […]

2. […] This approach can be extended to the complex series, which is looked at in the blog  God Plays Dice […]