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Question Number 38109 by maxmathsup by imad last updated on 21/Jun/18 | ||
$$\left.\mathrm{1}\right)\:{find}\:\:{S}\left({x}\right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{cos}\left({nx}\right)}{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}} \\ $$$$ \\ $$ | ||
Commented by abdo.msup.com last updated on 30/Jun/18 | ||
$${we}\:{have}\:{S}\left({x}\right)={Re}\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{inx}} }{{n}}\right)={Re}\left({w}\left({x}\right)\right) \\ $$$${w}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{inx}} }{{n}}\:\Rightarrow{w}^{'} \left({x}\right)={i}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left({e}^{{ix}} \right)^{{n}} \\ $$$$={i}\:{e}^{{ix}} \:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left({e}^{{ix}} \right)^{{n}−\mathrm{1}} ={ie}^{{ix}} \sum_{{n}=\mathrm{0}} ^{\infty} \:\left({e}^{{ix}} \right)^{{n}} \\ $$$$={i}\frac{{e}^{{ix}} }{\mathrm{1}−{e}^{{ix}} }\:={i}\:\frac{{e}^{{ix}} }{\mathrm{1}−{cosx}\:−{isinx}} \\ $$$$={i}\:\frac{{e}^{{ix}} }{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{2}{isin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$$=\:\frac{{i}}{−\mathrm{2}{isin}\left(\frac{{x}}{\mathrm{2}}\right)}\:\frac{{e}^{{ix}} }{{e}^{{i}\frac{{x}}{\mathrm{2}}} } \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right)}\:{e}^{{i}\frac{{x}}{\mathrm{2}}} \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)}\left\{\:{cos}\left(\frac{{x}}{\mathrm{2}}\right)\:+{i}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right)\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{cotan}\left(\frac{{x}}{\mathrm{2}}\right)\:−\frac{{i}}{\mathrm{2}}\:\Rightarrow \\ $$$${w}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{{cos}\left(\frac{{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}\:−\frac{{ix}}{\mathrm{2}}\:+{c} \\ $$$$=−{ln}\mid{sin}\left(\frac{{x}}{\mathrm{2}}\right)\mid\:−\frac{{ix}}{\mathrm{2}}\:+{c}\: \\ $$$${w}\left(\pi\right)\:=−\frac{{i}\pi}{\mathrm{2}}\:+{c}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right) \\ $$$$\Rightarrow{c}=\frac{{i}\pi}{\mathrm{2}}\:−{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$${w}\left({x}\right)=−{ln}\mid\:{sin}\left(\frac{{x}}{\mathrm{2}}\right)\mid−\frac{{ix}}{\mathrm{2}}\:+\frac{{i}\pi}{\mathrm{2}}\:−{ln}\left(\mathrm{2}\right) \\ $$$${S}\left({x}\right)={Re}\left({w}\left({x}\right)\right)=−{ln}\mid{sin}\left(\frac{{x}}{\mathrm{2}}\right)\mid−{ln}\left(\mathrm{2}\right). \\ $$$${also}\:{we}\:{get}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({nx}\right)}{{n}}\:=\frac{\pi−{x}}{\mathrm{2}}\:. \\ $$ | ||
Commented by abdo.msup.com last updated on 30/Jun/18 | ||
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{provef}\:{that} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{x}^{{n}} \:={ln}\left(\mathrm{1}+{x}\right)\:{if}\:\mid{x}\mid\leqslant\mathrm{1}\:{and} \\ $$$${x}\neq−\mathrm{1}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:={ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right). \\ $$ | ||