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In analyzing the convergence of the following series:

$$\sum_{n=1}^{\infty} \frac{n^n}{(n+1)!}$$

Using the quotient criteria, I get the following:

$$ \frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)^{n+1}}{(n+2)!}}{\frac{n^n}{(n+1)!}} = \frac{(n+1)^{n+1}(n+1)!}{(n+2)!n^n} = \frac{(n+1)^{n+1}(n+1)!}{(n+2)(n+1)!n^n} = \frac{(n+1)^{n+1}}{(n+2)n^n} $$

Now I am trying to find the limit of $\frac{a_{n+1}}{a_n}$:

$$ \lim_{n \to \infty}\frac{a_{n+1}}{a_n} = \lim_{n \to \infty}\frac{(n+1)^{n+1}}{(n+2)n^n} = ? $$

But I am having trouble solving this. What steps do I need to follow to simplify and solve this limit?

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    $\begingroup$ Rewrite it as$$\frac{a_{n+1}}{a_n}=\frac{n+1}{n+2}\cdot\left(1+\frac1n\right)^n$$ $\endgroup$ – Simply Beautiful Art Dec 14 '17 at 15:06
  • $\begingroup$ You need not apply the ratio test since the general terms don't approach $0$. In fact, $a_n\to \infty$. $\endgroup$ – Mark Viola Dec 14 '17 at 15:33
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Hint:

$$\frac{(n+1)^{n+1}}{n^n}=(n+1)\left(1+\frac1n\right)^n$$

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