Question Number 27912 by abdo imad last updated on 16/Jan/18
$${find}\:{the}\:{sum}\:{of}\:\sum_{{n}=\mathrm{1}} ^{\propto} \frac{\mathrm{1}}{{n}}\left(\:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{1}+{i}}\right)^{{n}} . \\ $$
Commented by abdo imad last updated on 23/Jan/18
$${for}\:{z}\in\:{iR}\:\:{and}\:\:\mid{z}\mid\leqslant\:\mathrm{1}\:{we}\:{have}\:\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\frac{{z}^{{n}} }{{n}}=\:−{ln}\left(\mathrm{1}−{z}\right)\: \\ $$$${but}\:{we}\:{have}\:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{1}+{i}}=\:\frac{\mathrm{1}}{{e}^{{i}\frac{\pi}{\mathrm{4}}} }\:={e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:{with}\:\mid\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \mid=\mathrm{1}\:\:{so} \\ $$$$\Sigma\:\frac{\mathrm{1}}{{n}}\:\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{1}+{i}}\right)^{{n}} =−{ln}\left(\mathrm{1}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:\:. \\ $$
Commented by abdo imad last updated on 23/Jan/18
$${ln}\:{here}\:{is}\:{a}\:{complex}\:{logarithme}. \\ $$