Question Number 102686 by Dwaipayan Shikari last updated on 10/Jul/20
$$\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}{i}^{{r}} +\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{i}^{{r}} \\ $$
Commented by Dwaipayan Shikari last updated on 10/Jul/20
$$\left({i}+{i}^{\mathrm{2}} +{i}^{\mathrm{3}} +{i}^{\mathrm{4}} +…..\right)=\frac{{i}}{\mathrm{1}−{i}} \\ $$$$\left(\mathrm{1}+{i}+{i}^{\mathrm{2}} +{i}^{\mathrm{3}} +……\right)=\frac{\mathrm{1}}{\mathrm{1}−{i}} \\ $$$$\frac{{i}}{\mathrm{1}−{i}}+\frac{\mathrm{1}}{\mathrm{1}−{i}}={i} \\ $$