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Well, the sequence is:

$a_n = n! - n^n$

I can't seem to figure it out. $n!$ goes to infinity, $n^n$ goes to infinity, I know the the result should be negative infinity, but I can't really find a way to explain it.

Just to note: I am embarrassed to even ask that question.

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Hint: $\dfrac{n^n}{n!} \ge n$.

Added: Let $n \ge 2$. The number $n!$ is the product of the $n-1$ terms $n, n-1, n-2, \dots, 2$, all of which are $\le n$. Thus $n! \le n^{n-1}$, and therefore $\dfrac{n^n}{n!} \ge n$.

Now $a_n=n!\left(1-\frac{n^n}{n!}\right) \le 1-n.$ But $\displaystyle \lim_{n\to \infty}(1-n) =-\infty$. Since $a_n\le 1-n$, it follows that $\displaystyle \lim_{n\to \infty}(n!-n^n) =-\infty$.

Comment: In situations of this type, some people prefer to say that the limit does not exist. That assertion is less informative, since it tells us much less about the behaviour of $a_n$ as $n$ gets very large.

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    No problem! I will put more detail in the main post.2011-11-28
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HINT: write $a_n = n^n(n!/n^n - 1)$ and try to prove that the sequence $b_n = n!/n^n$ converges to zero.