Discuss the convergence of the sequence whose $n$-th term is given by $a_n= \frac{n!}{n^n}$ Discuss the convergence of the sequence whose $n$-th term is given by
$$a_n= \frac{n!}{n^n}$$
I'm just a little confuse on how this can be solved because the factorial is confusing me.
 A: Put
$$a_n:=\frac{n!}{n^n}\implies\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e<1$$
thus the infinite positive series $\;\sum a_n\;$  converges, so $\;a_n\xrightarrow[n\to\infty]{}\;?$
A: Hint: Write out the factorial and $n^n$ then split it into $n$ fractions.
A: The ratio test is a great way to solve this. Examine the quantity $$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n} \right|$$ You can drop the absolute value since your sequence is strictly positive. Then you'll end up with $$\lim_{n \to \infty} \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} = \lim_{n \to \infty} \frac{(n+1)!n^n}{n!(n+1)^{n+1}}$$ You said factorials throw you off, but this next step is important; I remember it confusing people in my calculus class. By definition, $$(n+1)! = (n+1)\cdot n \cdot (n-1) \cdot \dots \cdot 2 \cdot 1 \\ = (n+1)\left[n \cdot (n-1) \cdot \dots \cdot 2 \cdot 1  \right] \\ = (n+1)[n!]$$ Now let's plug this rewritten version of $(n+1)!$ back into our limit to get $$\lim_{n \to \infty} \frac{(n+1)!n^n}{n!(n+1)^{n+1}} = \lim_{n \to \infty} \frac{(n+1)\cdot n!n^n}{n!\cdot(n+1)^{n+1}} \\ =\lim_{n \to \infty} \frac{n!}{n!} \frac{(n+1)}{(n+1)^{n+1}}
n^n$$ I broke up the fraction in the last line so it'd be easy to see what will cancel. When all is said and done, all you will have left is $$\lim_{n \to \infty} \frac{n^n}{(n+1)^{n}} = \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^n $$ I'll leave the rest of the limit to you. I'd guess L'Hospital's rule is the best tool you have for this. Once you show $\lim_{n \to \infty} \left(\frac{n}{n+1}\right)^n <1$ You will have shown that your sequence converges. Alternatively, you can also observe that one definition of $e$ is $$e = \lim_{n \to \infty} \left(1+\frac{1}{n} \right)^n \\ = \lim_{n \to \infty} \left(\frac{n+1}{n} \right)^n$$ which means $$\frac{1}{e} = \lim_{n \to \infty} \frac{1}{\left(\frac{n+1}{n} \right)^n} \\ =\lim_{n \to \infty} \left(\frac{n+1}{n} \right)^{-n} \\ = \lim_{n \to \infty} \left[\left(\frac{n+1}{n} \right)^{-1}\right]^n \\ = \lim_{n \to \infty} \left(\frac{n}{n+1} \right)^n$$ where the last equality is exactly the limit you were interested in evaluating.
A: $$a_n  = \frac{{n!}}{{n^n }} \cong \frac{{\sqrt {2\pi n} }}{{e^n }} \Rightarrow \mathop {\lim }\limits_{n \to  + \infty } a_n  \cong \mathop {\lim }\limits_{n \to  + \infty } \frac{{\sqrt {2\pi n} }}{{e^n }} = 0
$$
