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

Question Number 62656 by mathmax by abdo last updated on 24/Jun/19

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

Commented by mathmax by abdo last updated on 24/Jun/19

l e t A_{n} = \frac{{(n + 1)}^{n}}{n^{n + 1}} \Rightarrow A_{n} = {(\frac{n + 1}{n})}^{n} \times \frac{1}{n} = {(1 + \frac{1}{n})}^{n} . \frac{1}{n}

{(1 + \frac{1}{n})}^{n} = e^{n l n (1 + \frac{1}{n})} \to e (n \to + \infty) \Rightarrow {l i m}_{n \to + \infty} A_{n} = {l i m}_{n \to + \infty} \frac{e}{n} = 0

Answered by MJS last updated on 24/Jun/19

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

\lim_{n \to + \infty} \frac{n^{n - k}}{n^{n + 1}} = 0 for 0 ⩽ k ⩽ n \Rightarrow \lim_{n \to + \infty} \frac{{(n + 1)}^{n}}{n^{n + 1}} = 0

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