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Question Number 33891 by math khazana by abdo last updated on 26/Apr/18

$$leta\in Cand\mid a\mid <1provethatthefunction$$$$f\left(x\right)=\sum _{n=0}^{+\mathrm{\infty }}\frac{a^n}{x+n}is?developpableatpoint1and$$$$theradiusisr=1.$$

Commented by abdo imad last updated on 28/Apr/18

$$fis{C}^{\mathrm{\infty }}insideRand{f}^{\left(p\right)}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{a}^n\frac{{(-1)}^{p}p!}{{(x+n)}^{p+1}}$$$$\Rightarrow f^{\left(p\right)}\left(1\right)=\sum _{n=0}^{\mathrm{\infty }}{a}^n\frac{{(-1)}^{p}p!}{{(n+1)}^{p+1}}={(-1)}^{p}p!\sum _{n=0}^{\mathrm{\infty }}\frac{a^n}{{(n+1)}^{p+1}}$$$$={(-1)}^{p}p!\sum _{n=1}^{\mathrm{\infty }}\frac{a^{n-1}}{n^{p+1}}andwehave$$$$f\left(x\right)=\sum _{p=0}^{\mathrm{\infty }}\frac{f^{\left(p\right)}\left(1\right)}{p!}{(x-1)}^p$$$$=\sum _{p=0}^{\mathrm{\infty }}{(-1)}^p\left(\sum _{n=1}^{\mathrm{\infty }}\frac{a^{n-1}}{n^{p+1}}\right){(x-1)}^p.$$$$$$

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