91Ó°ÊÓ

The given series is$\underset{k=1}{\overset{∞}{∑}}\frac{k!}{k^k}.$

To show that the given series converges we will use the ratio test.

Let the general term is $a_k=\frac{k!}{k^k}.$

So, $a_{k+1}=\frac{\left(k+1\right)!}{{\left(k+1\right)}^{k+1}}.$

Now,

$$\begin{gathered} \mathrm{ÒÏ}=\underset{k→∞}{\lim }\frac{a_{k+1}}{a_k} \\ \mathrm{ÒÏ}=\underset{k→∞}{\lim }\frac{{\displaystyle \frac{\left(k+1\right)!}{{\left(k+1\right)}^{k+1}}}}{{\displaystyle \frac{k!}{k^k}}} \\ \mathrm{ÒÏ}=\underset{k→∞}{\lim }\frac{\left(k+1\right)!k^k}{{\left(k+1\right)}^{k+1}k!} \\ \mathrm{ÒÏ}=\underset{k→∞}{\lim }\frac{\left(k+1\right)k!\left(k^k\right)}{{(k+1)}^k(k+1)k!} \\ \mathrm{ÒÏ}=\underset{k→∞}{\lim }\frac{k^k}{{\left(k+1\right)}^k} \\ \mathrm{ÒÏ}=\frac{1}{e} \end{gathered}$$

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

To prove that $\underset{k→∞}{\lim }\frac{k!}{k^k}=0$we will use part (a), as we have shown in part (a) that $\underset{k=1}{\overset{∞}{∑}}\frac{k!}{k^k}$converges, so if the series converges then$\underset{k→∞}{\lim }\frac{k!}{k^k}=0.$

Hence proved.

To prove that part (b) proves that the function $k^k$dominates factorial growth.

Take the series $\underset{k=1}{\overset{∞}{∑}}\frac{k!}{k^k}.$

Now, as $k→∞$the denominator $k^k$will increase more than $k!.$

Thus, by the definition of dominance, the function $k^k$dominates factorial growth.