developp-f-x-e-x-x-1-at-integr-serie-

Question Number 33293 by abdo imad last updated on 14/Apr/18

d e v e l o p p f (x) = \frac{e^{x}}{x - 1} a t i n t e g r s e r i e

Commented by abdo imad last updated on 19/Apr/18

- f (x) = \frac{e^{x}}{1 - x} s o i f ∣ x ∣< 1 w e h a v e

- f (x) = (\sum_{n = 0}^{\infty} \frac{x^{n}}{n!}) (\sum_{n = 0}^{\infty} x^{n}) = (\sum_{n = 0}^{\infty} a_{n} x^{n}) (\sum_{n = 0}^{\infty} b_{n} x^{n})

= \sum_{n = 0}^{\infty} c_{n} x^{n} w i t h c_{n} = \sum_{i + j = n} a_{i} b_{j} = \sum_{i = 0}^{n} a_{i} b_{n - i}

= \sum_{i = 0}^{n} \frac{1}{i!} \Rightarrow f (x) = - \sum_{n = 0}^{\infty} (\sum_{i = 0}^{n} \frac{1}{i!}) x^{n} .