Q. 61 - Calculus

Question

Let n be a positive integer and let r > 1.

(a) Show that the series $\sum _{k=1}^{\infty }\frac{k^n}{r^k}$ converges.

(b) Explain why part (a) proves that$\underset{k\to \infty }{\lim }\frac{k^n}{r^k}=0.$

(c) Explain why part (b) proves that exponential growth dominates polynomial growth.

Step-by-Step Solution

Answer

Part a. The given series converges.

Part b. It proved that$\underset{k\to \infty }{\lim }\frac{k^n}{r^k}=0.$

Part c. Exponential growth dominates polynomial growth because as $k\to \infty$ the polynomial function $k^n$ will increase more than exponential growth $r^k.$ Thus, by the definition of dominance exponential growth dominates polynomial growth.

Part (a) Step 1. Given Information.

The given series is $\sum _{k=1}^{\infty }\frac{k^n}{r^k}.$

Part (a) Step 2. Showing that the given series converges.

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,

$$\begin{gathered} \rho =\underset{k\to \infty }{\lim }{\left(\frac{k^n}{r^k}\right)}^{\frac{1}{k}} \\ \rho =\underset{k\to \infty }{\lim }\left(\frac{k^{n\left({\displaystyle \frac{1}{k}}\right)}}{r^{k\left({\displaystyle \frac{1}{k}}\right)}}\right) \\ \rho =\underset{k\to \infty }{\lim }\left(\frac{k^{n\left({\displaystyle \frac{1}{k}}\right)}}{r^{k\left({\displaystyle \frac{1}{k}}\right)}}\right) \\ \rho =\underset{k\to \infty }{\lim }\frac{{\left(k^{\left({\displaystyle \frac{1}{k}}\right)}\right)}^n}{r} \\ \rho =\frac{1}{r} \end{gathered}$$

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

Thus, $\rho <1,$ by the root test, the given series converges.

Hence proved.

Part (b) Step 1. Explaining.

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

Hence proved.

Part (c) Step 1. Explaining.

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

Now, as $k\to \infty ,$ 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.