Question

For which nonnegative integer’s n is $n^2\le n!?$Prove your answer.

Short answer

P(n) is true for all integers n>3.

Step-by-step solution

Step: 1

The intersection of these two graphs is (3.5624, 12.6906).

Since n is an integer :n>3.

If n=4,

$$\begin{gathered} 4^2\le 4! \\ 16<24 \end{gathered}$$

it is true for n=4.

Step: 2

Let P(k) be true.

$k^2\le k!$

We need to prove that P(k+1) is true.

Step: 3

$\begin{array}{r}(k+1{)}^2=k^2+2k+\\ \le k!+2k+1\\ =k!+k\left(2+\frac{1}{k}\right)\\ =k!+kk!\\ =(k+1)k!\end{array}$

It is true for P(k+1) is true.