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Question Number 111982 by houssam last updated on 05/Sep/20

Answered by aleks041103 last updated on 05/Sep/20

$$\underset{k=0}{\sum ^{\mathrm{\infty }}}{a}^k=\frac{1}{1-a},for\mid a\mid <1$$$$Butherea=i,forwhich\mid a\mid =\mid i\mid =1,which$$$$meansthatthesum{doesn}^{\prime }tconverge$$

Answered by MJS_new last updated on 05/Sep/20

$$\mathrm{i}={\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}$$$${\mathrm{i}}^{4n}={\mathrm{e}}^{2\mathrm{i}\pi n}=1$$$${\mathrm{i}}^{4n+1}={\mathrm{e}}^{\mathrm{i}(\frac{\pi }{2}+2n\pi )}=\mathrm{i}$$$$\mathrm{similar}{\mathrm{i}}^{4n+2}=-1\wedge {\mathrm{i}}^{4n+3}=-\mathrm{i}$$$$\Rightarrow \underset{k=0}{\sum ^{\mathrm{\infty }}}{i}^k=1+\mathrm{i}-1-\mathrm{i}+1+...{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$

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