Chapter A2, Problem 6E

To determine

To show: $n!>n^3$ if $n$ is large enough.

Explanation of Solution

Proof:

Step 1: The inequality holds for $n=6$ because $6!=720>216=6^3$

Step 2: Suppose $k!>k^3$

Then $\left(k+1\right)!>\left(k+\right)k^3$

For $k\ge 4,k>1+\frac{2}{k}+\frac{1}{k^2}$

So, $k^3>k^2+2k+1={\left(k+1\right)}^2$

Then, $\left(k+1\right)k^3>{\left(k+1\right)}^3$ and by transitivity, $\left(k+1\right)!>{\left(k+1\right)}^3$

Hence, by principle of mathematical induction, this is true for all $n\ge 6$ . (Since the base case is true for $n=6$ , the relation is true only for $n$ greater than $6$ .)

Conclusion:

By principle of mathematical induction, this is true for all $n\ge 6$ .