Question Number 33891 by math khazana by abdo last updated on 26/Apr/18
$${let}\:{a}\in{C}\:{and}\:\mid{a}\mid<\mathrm{1}\:{prove}\:{that}\:{the}\:{function} \\ $$$${f}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{+\infty} \:\frac{{a}^{{n}} }{{x}+{n}}\:{is}?{developpable}\:{at}\:{point}\:\mathrm{1}\:{and} \\ $$$${the}\:{radius}\:{is}\:{r}=\mathrm{1}. \\ $$
Commented by abdo imad last updated on 28/Apr/18
$${f}\:{is}\:{C}^{\infty} \:{inside}\:{R}\:\:{and}\:{f}^{\left({p}\right)} \left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} {a}^{{n}} \frac{\left(−\mathrm{1}\right)^{{p}} {p}!}{\left({x}+{n}\right)^{{p}+\mathrm{1}} } \\ $$$$\Rightarrow\:{f}^{\left({p}\right)} \left(\mathrm{1}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{p}} {p}!}{\left({n}+\mathrm{1}\right)^{{p}+\mathrm{1}} }\:=\left(−\mathrm{1}\right)^{{p}} {p}!\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{a}^{{n}} }{\left({n}+\mathrm{1}\right)^{{p}+\mathrm{1}} } \\ $$$$=\left(−\mathrm{1}\right)^{{p}} {p}!\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{a}^{{n}−\mathrm{1}} }{{n}^{{p}+\mathrm{1}} }\:\:{and}\:{we}\:{have} \\ $$$${f}\left({x}\right)\:=\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{{f}^{\left({p}\right)} \left(\mathrm{1}\right)}{{p}!}\:\left({x}−\mathrm{1}\right)^{{p}} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{p}} \left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{a}^{{n}−\mathrm{1}} }{{n}^{{p}+\mathrm{1}} }\right)\left({x}−\mathrm{1}\right)^{{p}} \:\:. \\ $$$$ \\ $$