calculate-lim-n-n-1-1-n-n-en-

Question Number 63215 by mathmax by abdo last updated on 30/Jun/19

c a l c u l a t e {l i m}_{n \to + \infty} {n {(1 + \frac{1}{n})}^{n} - e n}

Commented by mathmax by abdo last updated on 01/Jul/19

l e t A_{n} = n {(1 + \frac{1}{n})}^{n} - e n \Rightarrow A_{n} = n {{(1 + \frac{1}{n})}^{n} - e} w e h a v e

{(1 + \frac{1}{n})}^{n} = e^{n l n (1 + \frac{1}{n})} b u t l n (1 + x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} x^{n}}{n} = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \dots \Rightarrow

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

e^{n l n (1 + \frac{1}{n})} = e^{1 - \frac{1}{2 n} + \frac{1}{3 n^{2}} o (\frac{1}{n^{2}})} = e e^{- \frac{1}{2 n} + \frac{1}{3 n^{2}} + o (\frac{1}{n})} d u e t o e^{u} = 1 + u + o (u) \Rightarrow

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

A_{n} = \frac{e}{2} + \frac{e}{3 n} + o (\frac{1}{n}) \Rightarrow {l i m}_{n \to + \infty} A_{n} = \frac{e}{2}

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