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

$$calculate{lim}_{n\to +\mathrm{\infty }}\frac{{(n+1)}^n}{n^{n+1}}$$

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

$$let{A}_n=\frac{{(n+1)}^n}{n^{n+1}}\Rightarrow A_n={\left(\frac{n+1}{n}\right)}^n\times \frac{1}{n}={(1+\frac{1}{n})}^n.\frac{1}{n}$$$${(1+\frac{1}{n})}^n=e^{nln(1+\frac{1}{n})}\to e(n\to +\mathrm{\infty })\Rightarrow {lim}_{n\to +\mathrm{\infty }}{A}_n={lim}_{n\to +\mathrm{\infty }}\frac{e}{n}=0$$

Answered by MJS last updated on 24/Jun/19

$$\frac{{(n+1)}^n}{n^{n+1}}=\frac{n^n}{n^{n+1}}+c_1\frac{n^{n-1}}{n^{n+1}}+c_2\frac{n^{n-2}}{n^{n+1}}+...+c_n\frac{1}{n^{n+1}}$$$$\underset{n\to +\mathrm{\infty }}{\lim }\frac{n^{n-k}}{n^{n+1}}=0\mathrm{for}0⩽k⩽n\Rightarrow \underset{n\to +\mathrm{\infty }}{\lim }\frac{{(n+1)}^n}{n^{n+1}}=0$$

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