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I want to figure out whether this series converges or not: $$\sum_{n=1}^\infty \frac{(n+1)!}{n^n}$$

Calculating the limit from ratio test $\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n}$ I came to the $\lim\limits_{n \to \infty} \frac{(n+2)n^n}{(n+1)^{n+1}}$. The limit equals to $\frac{1}{e}$ as Wolfram solves it, however I am not sure how to solve it by hand.

Thanks in advance.

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$\lim\limits_{n \to \infty} \dfrac{(n+2)n^n}{(n+1)^{n+1}} = \lim\limits_{n \to \infty} \dfrac{(1+\frac{2}{n})\cdot(n^{n+1})}{(1+\frac{1}{n})^{n+1}\cdot (n^{n+1})} = \lim\limits_{n \to \infty} \dfrac{(1+\frac{2}{n})}{(1+\frac{1}{n})^{n+1}}$
The limit in the numerator is $1$ and the limit in the denominator is $e$

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$$\frac{(n+2)n^n}{(n+1)^{n+1}}=\frac{n+2}{n+1}\cdot\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}1\cdot\frac1e\ldots$$

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Another way to do this: For large $n,$

$$\frac{(n+1)!}{n^n} = (n+1)\cdot\frac{n}{n}\cdot \frac{n-1}{n}\cdot \frac{n-2}{n}\cdots \frac{3}{n}\cdot \frac{2}{n}\cdot \frac{1}{n} \le (n+1)\frac{6}{n^3}.$$

The last expression is on the order of $1/n^2$ as $n\to \infty,$ so our series converges

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