find-a-equivalent-of-A-n-0-n-1-1-x-n-n-dx-n-

Question Number 33340 by prof Abdo imad last updated on 14/Apr/18

f i n d a e q u i v a l e n t o f

A_{n} = \int_{0}^{n} \sqrt{1 + {(1 - \frac{x}{n})}^{n}} d x (n \to \infty)

Commented by prof Abdo imad last updated on 25/Apr/18

w e k n o w t h a t {(1 + u)}^{α} = 1 + α u + \frac{α (α - 1)}{2} α^{2} + o (α^{3})

{(1 - \frac{x}{n})}^{n} = 1 - x + \frac{n (n - 1)}{2} \frac{x^{2}}{n^{2}} + o (\frac{x^{3}}{n^{3}}) \Rightarrow

1 + {(1 - \frac{x}{n})}^{n} \sim 2 - x + \frac{n (n - 1) x^{2}}{2 n^{2}} (n \to + \infty) a n d

\sqrt{2 - x + \frac{n (n - 1)}{2 n^{2}} x^{2}} = \sqrt{2 (1 - \frac{x}{2} + \frac{n (n - 1) x^{2}}{4 n^{2}}}

\sim \sqrt{2} (1 - \frac{1}{2} (\frac{x}{2} - \frac{n (n - 1) x^{2}}{4 n^{2}}))

= \sqrt{2} - \frac{\sqrt{2}}{4} x + \frac{\sqrt{2}}{8} \frac{n (n - 1) x^{2}}{n^{2}} \Rightarrow

A_{n} \sim \int_{0}^{n} (\sqrt{2} - \frac{\sqrt{2}}{4} x + \frac{\sqrt{2}}{8} \frac{n (n - 1) x^{2}}{n^{2}}) d x \Rightarrow

A_{n} \sim n \sqrt{2} - \frac{\sqrt{2}}{8} n^{2} + \frac{\sqrt{2}}{24} \frac{n (n - 1) n^{3}}{n^{2}} \Rightarrow

A_{n} \sim n \sqrt{2} - \frac{\sqrt{2}}{8} n^{2} + \frac{\sqrt{2}}{24} n^{2} (n - 1) (n \to + \infty)