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Question Number 38725 by maxmathsup by imad last updated on 28/Jun/18
$${let}\:{f}\left({x}\right)={ln}\left(\mathrm{1}+\:{e}^{−{x}} \right)\:\:{developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$
Commented by abdo.msup.com last updated on 29/Jun/18
$${we}\:{have}\:{ln}\left(\mathrm{1}+{u}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{x}^{{n}} \Rightarrow \\ $$$${ln}\left(\mathrm{1}+{e}^{−{x}} \right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:{e}^{−{nx}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{\left(−{nx}\right)^{{p}} }{{p}!} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\left\{\sum_{{p}=\mathrm{0}} ^{\infty} \frac{\left(−{n}\right)^{{p}} }{{p}!}{x}^{{p}} \right\}. \\ $$$$ \\ $$