Question-121574

Question Number 121574 by oustmuchiya@gmail.com last updated on 09/Nov/20

Answered by TANMAY PANACEA last updated on 09/Nov/20

t = l i \underset{n \to \infty}{m} {(1 + \frac{1}{n})}^{n}

l n t = l i \underset{n \to \infty}{m n l n} (1 + \frac{1}{n})

l n t = l i \underset{y \to 0}{m} \frac{l n (1 + y)}{y} = 1

t = e^{1} = e

Answered by Dwaipayan Shikari last updated on 09/Nov/20

\lim_{n \to \infty} {(1 + \frac{1}{n})}^{n} = (1 + \frac{n}{n} + \frac{n (n - 1)}{2! n^{2}} + \frac{n (n - 1) (n - 2)}{3! n^{3}} + \dots .)

= (1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots .) = \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} = e

Answered by mathmax by abdo last updated on 09/Nov/20

u_{n} = {(1 + \frac{1}{n})}^{n} \Rightarrow u_{n} = e^{nln (1 + \frac{1}{n})} we have \ln (1 + u) = u - \frac{u^{2}}{2} + o (u^{3}) (u \sim 0)

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

u_{n} = e . e^{- \frac{1}{2 n} + o (\frac{1}{n^{2}})} \sim e {1 - \frac{1}{2 n}} \Rightarrow \lim_{n \to + \infty} u_{n} = e