let-a-C-and-a-lt-1-prove-that-the-function-f-x-n-0-a-n-x-n-is-developpable-at-point-1-and-the-radius-is-r-1-

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

l e t a \in C a n d ∣ a ∣< 1 p r o v e t h a t t h e f u n c t i o n

f (x) = \sum_{n = 0}^{+ \infty} \frac{a^{n}}{x + n} i s ? d e v e l o p p a b l e a t p o i n t 1 a n d

t h e r a d i u s i s r = 1 .

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

f i s C^{\infty} i n s i d e R a n d f^{(p)} (x) = \sum_{n = 0}^{\infty} a^{n} \frac{{(- 1)}^{p} p!}{{(x + n)}^{p + 1}}

\Rightarrow f^{(p)} (1) = \sum_{n = 0}^{\infty} a^{n} \frac{{(- 1)}^{p} p!}{{(n + 1)}^{p + 1}} = {(- 1)}^{p} p! \sum_{n = 0}^{\infty} \frac{a^{n}}{{(n + 1)}^{p + 1}}

= {(- 1)}^{p} p! \sum_{n = 1}^{\infty} \frac{a^{n - 1}}{n^{p + 1}} a n d w e h a v e

f (x) = \sum_{p = 0}^{\infty} \frac{f^{(p)} (1)}{p!} {(x - 1)}^{p}

= \sum_{p = 0}^{\infty} {(- 1)}^{p} (\sum_{n = 1}^{\infty} \frac{a^{n - 1}}{n^{p + 1}}) {(x - 1)}^{p} .