Sum of powers of i

James Tanton asks:

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

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