I am trying to check whether or not the sequence $$a_{n} =\left\{\frac{n^n}{n!}\right\}_{n=1}^{\infty}$$ is bounded, convergent and ultimately monotonic (there exists an $N$ such that for all $n\geq N$ the sequence is monotonically increasing or decreasing). However, I'm having a lot of trouble finding a solution that sufficiently satisfies me.

My best argument so far is as follows,

$$a_{n} = \frac{n\cdot n\cdot n\cdot \ldots\cdot n}{n(n-1)(n-2)(n-3)\dots(2)(1)} = \frac{n}{n}\cdot \frac{n}{n-1}\cdot \ldots \cdot \frac{n}2\cdot n$$

so $\lim a_{n}\rightarrow \infty$ since $n<a_{n}$ for all $n>1$. Since the sequence is divergent, it follows that the function must be ultimately monotonic.

This feels a little dubious to me, I feel like I can form a much better argument than that, or at the very least a more elegant one. I've tried to assume $\{a_{n}\}$ approaches some limit $L$ so there exists some $N$ such that

$|a_{n} - L| < \epsilon$ whenever $n>N$ and derive a contradiction, but this approach got me nowhere.

Finally, I've also tried to use the fact that $\frac{a_{n+1}}{a_n}\rightarrow e$ to help me, but I couldn't find an argument where that fact would be useful.

  • $\begingroup$ Your argument that $a_n\ge n$ is fine and shows that the sequence is unbounded and therefore not convergent; the only remaining question is whether it’s eventually monotonic. $\endgroup$ – Brian M. Scott Oct 22 '13 at 2:45
  • $\begingroup$ So we are talking about a sequence here. The question is. What grows faster: Numerator or denominator? Suggestion: Plug in couple of values of n, and see what happens... $\endgroup$ – imranfat Oct 22 '13 at 2:46
  • $\begingroup$ I suggest you have a look to Stirling's approximation of n!. Using it, you arrive to the fact that n^n/n! is almost equal to Exp[n]/Sqrt[2 Pi n]. $\endgroup$ – Claude Leibovici Oct 22 '13 at 5:59
  • $\begingroup$ Similar posts: math.stackexchange.com/questions/61713/… and math.stackexchange.com/questions/397866/… $\endgroup$ – Martin Sleziak Dec 17 '13 at 15:34

HINT for the last part: Note that


  • $\begingroup$ Using this, I can show that given $\epsilon>0$, there exists an $N$ such that for all $n> N$, $a_{n} (e-\epsilon) <a_{n+1}$, but I can't seem to reduce it further. I think I am missing a well known theorem about the ratio for terms that is preventing me from completing this problem. $\endgroup$ – JessicaK Oct 22 '13 at 2:59
  • 1
    $\begingroup$ @Jessica: You’re trying to be too fancy, I think. $$\left(\frac{n+1}n\right)^n=\left(1+\frac1n\right)^n\;;$$ how does that ratio compare in size with $1$? Remember, all you need to do is show that the sequence is monotonic; the rest is handled by your earlier work, as I said in a comment. $\endgroup$ – Brian M. Scott Oct 22 '13 at 3:05
  • $\begingroup$ I see what I'm doing wrong now. I was too obsessed with the limit that it did not occur to me that the ratio is greater than 1 before even applying the limit. Thank you very much. $\endgroup$ – JessicaK Oct 22 '13 at 3:12
  • $\begingroup$ @Jessica: You’re very welcome. $\endgroup$ – Brian M. Scott Oct 22 '13 at 3:13

To show that $a_n$ is unbounded, we can show that, $a_n > n$ for all $n$. And since $n$ is unbounded by Archimedes, $a_n$ must also be unbounded.

$a_n = \frac{n^n}{n!} = \underbrace{\frac{n}{n} \cdot \frac{n}{n-1} \cdot \frac{n}{n-2} \cdot \frac{n}{n-3}\cdots \frac{n}{2}}_{\text{Since $n \geq 1$, each of these terms are $\geq 1$}} \cdot \frac{n}{1} \geq n$.


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