Proving n! > n2 for n ≥ 4 and n! > n3 for n ≥ 6 Using Mathematical Induction

One common approach to proving the inequality n! > n2 for n ≥ 4 and n! > n3 for n ≥ 6 is by utilizing mathematical induction. This method is particularly useful in verifying such statements involving sequences and series. Unlike the use of Stirling's approximation, this technique relies on the step-by-step verification through base cases and inductive steps. Let's delve into the detailed proof for each inequality.

Proof for n! > n2 for n ≥ 4

First, we establish the base case:

4! 24 16 42. This is clear, as we can directly calculate that 24 is greater than 16.

Next, for the inductive step, we assume that for some integer k ≥ 4, k! k2. We need to show that the inequality holds for (k 1)!. Starting with the inductive hypothesis:

(k 1) (k 1)k! (k 1)k2

We need to show that (k 1)k! (k 1)2. This simplifies to showing that k2 k. This is true for all k ≥ 4 because:

k2 - k k(k - 1) 0 for all k ≥ 4.

Therefore, by the principle of mathematical induction, n! n2 for all integers n ≥ 4.

Proof for n! > n3 for n ≥ 6

Similar to the previous proof, we begin with the base case:

6! 720 216 63. This is evident as 720 is clearly greater than 216.

For the inductive step, we assume that for some integer k ≥ 6, k! k3. We need to show that the inequality holds for (k 1)!. Starting with the inductive hypothesis:

(k 1) (k 1)k! (k 1)k3

We need to show that (k 1)k! (k 1)3. This simplifies to showing that k3 2k2 3k 1. This is true for all k ≥ 6 due to the following reasoning:

k3 - 3k2 - 3k - 1 0 for all k ≥ 6.

Therefore, by the principle of mathematical induction, n! n3 for all integers n ≥ 6.

Conclusion

In summary, we have proven that n! n2 for n ≥ 4 and n! n3 for n ≥ 6 using mathematical induction. These results are established by verifying both the base cases and the inductive steps, ensuring the validity of the inequalities for all specified integers.