deduce that limit equals 0

**twilightstr** · April 5th 2009, 08:29 PM

Deduce that lim as n goes to infinity of x^n/(n!) equals o for all x

**Chris L T521** · April 5th 2009, 08:48 PM

Originally Posted by twilightstr

Deduce that lim as n goes to infinity of x^n/(n!) equals o for all x

If $\lim\frac{s_{n+1}}{s_n}=L$ where $L<1$ , then $\lim s_n=0$

So, let $s_n=\frac{x^n}{n!}$ . As a result, $s_{n+1}=\frac{x^{n+1}}{\left(n+1\right)!}$

Therefore, $\lim\frac{s_{n+1}}{s_n}=\lim\frac{x^{n+1}}{\left(n +1\right)!}\cdot\frac{n!}{x^n}=\lim\frac{x}{n+1}=0 \quad\forall\,x\in\mathbb{R}$

Since $0<1$ , we can conclude that $\lim\frac{x^n}{n!}=0\quad\forall\,x\in\mathbb{R}$ .

(Note that $\lim$ is analogous with $\lim_{n\to\infty}$ )

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