Chapter 9.4, Problem 26E

To determine

To prove; the inequality for the specified integer values of $n$ .

Explanation of Solution

Given information:

Given inequality

  ${\left(\frac{4}{3}\right)}^n>n,n\ge 7$

Calculation:

1. For $n=7$ and $n=8$, the statement is true because

  $\begin{array}{c}{\left(\frac{4}{3}\right)}^7>7\\ 7.49>7\end{array}$

And

  $\begin{array}{c}{\left(\frac{4}{3}\right)}^8>8\\ 9.9>8\end{array}$

2. Now, assume that

  ${\left(\frac{4}{3}\right)}^k>k$

Now show that ${\left(\frac{4}{3}\right)}^{k+1}>k+1,k+1\ge 7$

For $n=k$ , we have

  $\begin{array}{c}{\left(\frac{4}{3}\right)}^{k+1}={\left(\frac{4}{3}\right)}^k\left(\frac{4}{3}\right)\\ =\frac{4k}{3}>k+1\end{array}$

Since, $\frac{4k}{3}=k+\frac{k}{3}>k+1$ for all $k+1\ge 7$ , it follows that

  ${\left(\frac{4}{3}\right)}^{k+1}>\frac{4k}{3}>k+1\text{or}k+1<{\left(\frac{4}{3}\right)}^{k+1}$

Combining the results of parts (1) and (2), we can conclude by mathematical induction that

  ${\left(\frac{4}{3}\right)}^n>n$ for all integers $n\ge 7$