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Question Number 49804 by maxmathsup by imad last updated on 10/Dec/18

l e t f (x) = \frac{e^{- x}}{x + 1}

1) c a l c u l a t e f^{(n)} (o) a n d f^{(n)} (1)

2) d e v e l o p p f a t i n t e g r s e r i e .

Commented by Abdo msup. last updated on 11/Dec/18

1) l e t f i r s t f i n d f^{(n)} (x)

f^{(n)} (x) = {(\frac{1}{x + 1} e^{- x})}^{(n)} l e i b n i z f o r m u l a g i v e

f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(\frac{1}{x + 1})}^{(k {} {(e^{- x})}^{(n - k)}

= \sum_{k = 0}^{n} C_{n}^{k} \frac{{(- 1)}^{k} k!}{{(x + 1)}^{k + 1}} {(- 1)}^{n - k} e^{- x}

= {(- 1)}^{n} \sum_{k = 0}^{n} \frac{k! C_{n}^{k}}{{(x + 1)}^{k + 1}} e^{- x}

= {(- 1)}^{n} \sum_{k = 0}^{n} \frac{n!}{(n - k)!} \frac{e^{- x}}{{(x + 1)}^{k + 1}} \Rightarrow

f^{(n)} (0) = n! {(- 1)}^{n} \sum_{k = 0}^{n} \frac{1}{(n - k)!} a n d

f^{(n)} (1) = n! {(- 1)}^{n} \sum_{k = 0}^{n} \frac{1}{(n - k)!} \frac{e^{- 1}}{2^{k + 1}} .

2) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}

= \sum_{n = 0}^{\infty} {(- 1)}^{n} (\sum_{k = 0}^{n} \frac{1}{(n - k)! :}) x^{n} b u t

\sum_{k = 0}^{n} \frac{1}{(n - k)!} =_{n - k = p} \sum_{p = 0}^{n} \frac{1}{p!} \Rightarrow

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