prove-that-1-lim-n-0-n-1-x-n-n-e-x-dx-1-1-

Question Number 33343 by prof Abdo imad last updated on 14/Apr/18

p r o v e t h a t \forall α \in] 1, + \infty [

{l i m}_{n \to \infty} \int_{0}^{n} {(1 + \frac{x}{n})}^{n} e^{- α x} d x = \frac{1}{α - 1} .

Commented by abdo imad last updated on 17/Apr/18

l e t p u t I_{n} = \int_{0}^{n} {(1 + \frac{x}{n})}^{n} e^{- α x} d x = \int_{R} f_{n} (x) d x w i t h

f_{n} (x) = {(1 + \frac{x}{n})}^{n} e^{- α x} χ_{[0, n [} (x) d x w e h a v e

f_{n} (x) \to^{c . s} f (x) = e^{(1 - α) x} i f x ⩾ 0 a n d f (x) = 0 i f x < 0

b u t w e h a v e {(1 + \frac{x}{n})}^{n} = e^{n l n (1 + \frac{x}{n})} b u t l n (1 + \frac{x}{n}) ⩽ \frac{x}{n} \Rightarrow

n l n (1 + \frac{x}{n}) ⩽ x \Rightarrow {(1 + \frac{x}{n})}^{n} e^{- α x} ⩽ g (x) =_{} e^{(1 - α) x}

t h e o r e m o f c o n v e r g e n c e d o m i n e e g i v e

\int_{R} f_{n} (x) d x_{n \to \infty} \to \int_{0}^{\infty} e^{(1 - α) x} d x = \frac{1}{1 - α} {[e^{(1 - α) x}]}_{0}^{+ \infty}

= \frac{- 1}{1 - α} = \frac{1}{α - 1} \Rightarrow {l i m}_{n \to \infty} I_{n} = \frac{1}{1 - α} .

Commented by prof Abdo imad last updated on 17/Apr/18

l i m I_{n} = \frac{1}{α - 1} .