Use the Ratio Test to determine whether the series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{k!}{k^{k}}$ converges or diverges.

Solution $\sum\limits_{k\,=\,1}^{\infty }\dfrac{k!}{k^{k}}$ is a series of nonzero terms. Since $a_{n+1}=\dfrac{(n+1)!}{(n+1) ^{n+1}}$ and $a_{n}=\dfrac{n!}{n^{n}}$, the absolute value of their ratio is $$\begin{eqnarray*} \begin{array}{@{\hspace*{-0pc}}rcl} \displaystyle\left\vert \dfrac{a_{n+1}}{a_{n}}\right\vert &=& \dfrac{\dfrac{(n+1)!}{(n+1) ^{n+1}}}{\dfrac{n!}{n^{n}}}=\dfrac{(n+1)!}{(n+1)^{n+1}}\cdot \dfrac{n^{n}}{n!}=\dfrac{n^{n}}{(n+1)^{n}}\\ &=&\left( \dfrac{n}{n+1}\right)^{n} = \left( \dfrac{1}{1+\dfrac{1}{n}}\right)^{n} = \dfrac{1}{\left( 1+\dfrac{1}{n}\right) ^{n}}\qquad \end{array} \end{eqnarray*}$$

So, $$\lim\limits_{n\,\rightarrow \,\infty }\left\vert \frac{a_{n+1}}{a_{n}} \right\vert =\lim\limits_{n\,\rightarrow \,\infty }\frac{1}{\left( 1+\dfrac{1 }{n}\right) ^{n}}=\frac{\lim\limits_{n\,\rightarrow \,\infty }1}{ \lim\limits_{n\,\rightarrow \,\infty }\,\left( 1+\dfrac{1}{n}\right) ^{n}}=\frac{1}{e}$$

Since $\dfrac{1}{e}<1$, the series converges.