Use the ratio or the root test to show convergence or divergence of the following series. If inconclusive, use another test:
$$\sum_{n=1}^{\infty}\frac{n!}{n^{n}}$$
So my first instinct was to try the ratio test due to the existence of the factorial. This is my working:
Using the Ratio Test:
\begin{align*}
\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|&=\lim_{n\to\infty}\left|\frac{(n+1)\cdot n!\cdot n^{n}}{n!\cdot n\cdot n^{n}}\right|\\
&=\lim_{n\to\infty}\left|\frac{n+1}{n}\right|\\
&=1
\end{align*}
The Ratio Test is inconclusive.
I decided then to try the root test due to the presence of the $n^n$, but I think that's problematic and won't work (unless I'm looking at something the wrong way). I end up with the following:
\begin{align}
\lim_{n\to\infty}\left|a_n\right|^{1/n}&=\lim_{n\to\infty}\left|\frac{n!}{n^n}\right|\\
&=\lim_{n\to\infty}\left|\frac{(n!)^{1/n}}{n}\right|=\frac{\infty}{\infty}
\end{align}
So my problem here is that I can't apply L'Hopital's rule. If I expand the numerator I get the following:
$$(n!)^{1/n}=\sqrt[n]{n}\cdot\sqrt[n]{n-1}\cdot\sqrt[n]{n-2}\cdot\sqrt[n]{n-3}\cdots\sqrt[n]{3}\cdot\sqrt[n]{2}\cdot\sqrt[n]{1}$$
Which then would only allow me to cancel the $\sqrt[n]{n}$ and get $n^{(n-1)/n}$ in the denominator. Still gives me the indeterminate form of $\infty/\infty$.
So how can I approach this? Or was I on the right track and did something wrong?
Best Answer
$$ \frac{n^n}{(n+1)^{n+1}} = \frac{1}{n+1}\cdot\frac{n^n}{(n+1)^n}. $$ The $n+1$ in the denominator is canceled by the $n+1$ in the other part of the whole expression. Then we have $$ \frac{n^n}{(n+1)^n}\to \frac1e\text{ as }n\to\infty. $$