# $\lim\limits_{n\to\infty}\frac{n^{n+1}}{n!}$ with sandwich rule

Determine the result of

$$\lim\limits_{n\to\infty}\frac{n^{n+1}}{n!}.$$

I would like to use the sandwich rule for limits to find two sequences which define lower and upper bounds to determine the appropriate limit which is obviously $+\infty$.

By now I have an upper bound:

$$\frac{n^{n+1}}{n!}=n\cdot\frac{n}{n}\cdot\frac{n}{n-1}\cdot\ldots\cdot\frac{n}{2}\cdot\frac{n}{1}\leq n\cdot 1\cdot n\cdot\ldots\cdot n\cdot n=n^n\longrightarrow+\infty$$

What would you suggest, which sequence should I use for a lower bound which converges to $+\infty$?

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$$\frac{n^{n+1}}{n!}=n\frac{n\cdot n\cdot\ldots\cdot n}{1\cdot 2\cdot\ldots\cdot n}\geq n\frac{n\cdot n\cdot\ldots\cdot n}{n\cdot n\cdot\ldots\cdot n}=n\xrightarrow [n\to\infty]{} \infty$$
I'd take out $\,\dfrac{n}{1}\,$ from the fraction: $$\frac{n^n}{n!}\geq \frac{n}{1}\frac{n\cdot...\cdot n}{n\cdot\ldots\cdot n}=n....$$ –  DonAntonio Nov 5 '12 at 14:05