$\displaystyle \sum_{n=1}^{\infty}\frac{n!}{n^n}$
determine whether the series converges or diverges
help?
Hello, buttonbear!
Without the Ratio Test, I'm not sure of the proof.
The Ratio Test provides some interesting maniputations . . .$\displaystyle \sum_{n=1}^{\infty}\frac{n!}{n^n}$ . Determine whether the series converges or diverges
$\displaystyle R \;=\;\frac{a_{n+1}}{a_n} \;=\;\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!} \;=\;\frac{(n+1)!}{n!}\cdot\frac{n^n}{(n+1)^{n+1}}$
. . $\displaystyle = \;\frac{n+1}{1}\cdot\frac{n^n}{(n+1)^{n+1}} \;=\;\frac{n^n}{(n+1)^n} \;=\;\left(\frac{n}{n+1}\right)^n$
Divide top and bottom of the fraction by $\displaystyle n\!:\;\;\left(\frac{1}{1+\frac{1}{n}}\right)^n \;=\;\frac{1}{\left(1 + \frac{1}{n}\right)^n} $
Therefore: .$\displaystyle \lim_{n\to\infty}R \;=\;\lim_{n\to\infty}\frac{1}{\left(1 + \frac{1}{n}\right)^n} \;=\;\frac{1}{\lim\left(1 + \frac{1}{n}\right)^n} \;=\;\frac{1}{e} \;<\;1$
. . and the series converges.
For all $\displaystyle n \geq 2$, we have: $\displaystyle \frac{n!}{n^n} = \frac{1 \cdot 2 \cdot 3 \cdots (n-1)n}{n \cdot n \cdot n \cdots n \cdot n} \ \ {\color{red} \leq}\ \ \frac{1}{n} \cdot \frac{2}{n} \cdot 1 \cdot 1 \cdots \cdot 1 = \frac{2}{n^2}$
So by comparison ...