# Why is the sequence divergent?

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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Just to be sure, you mean the comparison test, right? Does that apply to sequences? –  PawelMysior Nov 28 '11 at 14:48
A version of comparison does apply to sequences, but it is best not to think that way. We have $n!-n^n=n!(1-\frac{n^n}{n!})\le n!(1-n)$ (the last is by the Hint). But $\displaystyle \lim_{n\to\infty}n!(1-n)=-\infty$ (or doesn't exist, it depends what conventions are used in your course). So depending on those conventions, you can say your original limit is $-\infty$, or that it doesn't exist. –  André Nicolas Nov 28 '11 at 15:45
Sorry to be bugging You again, but I can't really get the grip on your theory. I get that the smaller sequence is divergent to negative infinity, but why is the bigger one, I think I'm missing some basic knowledge here... could You please point me in the right direction? –  PawelMysior Nov 28 '11 at 17:26
No problem! I will put more detail in the main post. –  André Nicolas Nov 28 '11 at 18:07

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.

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