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Question Number 33846 by prof Abdo imad last updated on 26/Apr/18
$${find}\:{radous}\:{of}\:{conbergence}\:{for}\:{theserie}\:\sum_{{n}\geqslant\mathrm{0}} {x}^{{n}!} .\: \\ $$
Commented by prof Abdo imad last updated on 31/May/18
$$\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}!} \:{is}\:{a}\:{extract}\:{serie}\:{from}\:{the}\:{serie} \\ $$$$\sum_{{n}=\mathrm{0}} ^{+\infty} \:{x}^{{n}} \:\:{wich}\:{converges}\:{if}\:\mid{x}\mid<\mathrm{1}\:\:{so}\:{R}\:\leqslant\mathrm{1}\:. \\ $$