91Ó°ÊÓ

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

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

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

So,

$$\begin{gathered} \mathrm{ÒÏ}=\underset{k→∞}{\lim }{\left(\frac{k^n}{r^k}\right)}^{\frac{1}{k}} \\ \mathrm{ÒÏ}=\underset{k→∞}{\lim }\left(\frac{k^{n\left({\displaystyle \frac{1}{k}}\right)}}{r^{k\left({\displaystyle \frac{1}{k}}\right)}}\right) \\ \mathrm{ÒÏ}=\underset{k→∞}{\lim }\left(\frac{k^{n\left({\displaystyle \frac{1}{k}}\right)}}{r^{k\left({\displaystyle \frac{1}{k}}\right)}}\right) \\ \mathrm{ÒÏ}=\underset{k→∞}{\lim }\frac{{\left(k^{\left({\displaystyle \frac{1}{k}}\right)}\right)}^n}{r} \\ \mathrm{ÒÏ}=\frac{1}{r} \end{gathered}$$

As it is given that r > 1,so $\frac{1}{r}<1.$

Thus, $\mathrm{ÒÏ}<1,$by the root test, the given series converges.

Hence proved.

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

Hence proved.

To prove that part (b) proves that exponential growth dominates polynomial growth. We will let the polynomial function as $k^n.$

Now, as $k→∞,$the polynomial function $k^n$will increase more than exponential growth $r^k.$

Thus, by the definition of dominance, exponential growth dominates polynomial growth.

Hence proved.