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Question Number 102686 by Dwaipayan Shikari last updated on 10/Jul/20

$$\underset{r=1}{\sum ^{\mathrm{\infty }}}{i}^r+\underset{r=0}{\sum ^{\mathrm{\infty }}}{i}^r$$

Commented by Dwaipayan Shikari last updated on 10/Jul/20

$$(i+i^2+i^3+i^4+.....)=\frac{i}{1-i}$$$$(1+i+i^2+i^3+......)=\frac{1}{1-i}$$$$\frac{i}{1-i}+\frac{1}{1-i}=i$$

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