What is the sum from n=0 to infinity of (1-i)^n ?

The sum from n=0 to infinity of (1-i)^n is diverges

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sum from n=0 to infinity of (1-i)^n

n = 0 \sum \infty (1 - i)^{n}

diverges

Solution steps

n = 0 \sum \infty (1 - i)^{n}

Apply Series Ratio Test:

diverges

= diverges

n = 0 \sum \infty 5 \cdot (\frac{2}{3})^{n}

n = 1 \sum \infty \frac{1}{n ^{2 - 1}}

n = 600 \sum \infty \frac{1}{n}

n = 1 \sum \infty \frac{n}{e ^{7 n}}

n = 1 \sum \infty \frac{3 ^{n}}{6 ^{n}}

What is the sum from n=0 to infinity of (1-i)^n ?
The sum from n=0 to infinity of (1-i)^n is diverges