Help-me-obtain-the-value-of-e-from-1-1-n-n-how-do-i-go-about-it-

Question Number 67659 by Rio Michael last updated on 29/Aug/19

H e l p m e o b t a i n t h e v a l u e o f e f r o m

{(1 + \frac{1}{n})}^{n} h o w d o i g o a b o u t i t .

Commented by Abdo msup. last updated on 30/Aug/19

A_{n} = {(1 + \frac{1}{n})}^{n} a n d {l i m}_{n \to + \infty} A_{n} = e

A_{n} = \sum_{k = 0}^{n} C_{n}^{k} \frac{1}{n^{k}} = \frac{C_{n}^{0}}{n^{0}} + \frac{C_{n}^{1}}{n^{1}} + \frac{C_{n}^{2}}{n^{2}} + \frac{C_{n}^{3}}{n^{3}} + . . + \frac{C_{n}^{n}}{n^{n}} \Rightarrow

e = {l i m}_{n \to + \infty} A_{n} = 1 + 1 + \frac{C_{n}^{2}}{n^{2}} + \frac{C_{n}^{3}}{n^{3}} + \dots .

= 2 + \frac{n (n - 1)}{2 n^{2}} + \frac{n (n - 1) (n - 2)}{6 n^{3}} + \dots

w d f i n d e \sim 2, 7 \dots

Commented by Rio Michael last updated on 30/Aug/19

t h a n k s s i r

Commented by Abdo msup. last updated on 30/Aug/19

y o u a r e w e l c o m e .