Q. 63 - Calculus

Question

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

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

(c) Explain why part (b) proves that the function $k^k$ dominates factorial growth.

Step-by-Step Solution

Answer

Part a. The given series converges.

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

Part c. The function $k^k$ dominates factorial growth because as $k\to \infty$ the denominator $k^k$ will increase more than $k!.$ Thus, by the definition of dominance, the function $k^k$ dominates factorial growth.

Part (a) Step 1. Given Information.

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

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

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,

$$\begin{gathered} \rho =\underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k} \\ \rho =\underset{k\to \infty }{\lim }\frac{{\displaystyle \frac{\left(k+1\right)!}{{\left(k+1\right)}^{k+1}}}}{{\displaystyle \frac{k!}{k^k}}} \\ \rho =\underset{k\to \infty }{\lim }\frac{\left(k+1\right)!k^k}{{\left(k+1\right)}^{k+1}k!} \\ \rho =\underset{k\to \infty }{\lim }\frac{\left(k+1\right)k!\left(k^k\right)}{{(k+1)}^k(k+1)k!} \\ \rho =\underset{k\to \infty }{\lim }\frac{k^k}{{\left(k+1\right)}^k} \\ \rho =\frac{1}{e} \end{gathered}$$

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

Part (b) Step 1. Explaining.

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

Hence proved.

Part (c) Step 1. Explaining.

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

Take the series $\sum _{k=1}^{\infty }\frac{k!}{k^k}.$

Now, as $k\to \infty$ the denominator $k^k$ will increase more than $k!.$

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