r-1-i-r-r-0-i-r-

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

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

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

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

(1 + i + i^{2} + i^{3} + \dots \dots) = \frac{1}{1 - i}

\frac{i}{1 - i} + \frac{1}{1 - i} = i