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Question Number 35766 by abdo mathsup 649 cc last updated on 23/May/18
letf(x)=arctan(1+x)andfn(x)=e−nx1)definefofn(x)andfnof(x)2)studythenatureoftheseries∑n=0+∞fofn(x)and∑n=0∞fnof(x).
Commented by maxmathsup by imad last updated on 02/Sep/18
1)wehavefofn(x)=f(fn(x))=arctan(1+fn(x))=arctan(1+e−nx)fn0f(x)=fn(f(x))=e−nf(x)=e−narctan(1+x)2)∑n=0∞fofn(x)=∑n=0∞arctan(1+e−nx)=Σun(x)ifx<0limn→+∞arctan(1+e−nx)≠0soΣun(x)divergesifx>0un(x)∼e−nx(n→+∞)butΣe−nxconvergessotheconvergencesimpleofΣun(x)isassured.alsoΣfnof=∑ne−narctan(1+x)=∑n(e−arctan(1+x))n⇒ifarctan(1+x)>0theserieconvergesifarctan(1+x)<0theseriediverges.
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