The discussion confirms the validity of the inequality \(2^{n+2} < (n+1)!\) for all \(n \geq 6\). The proof utilizes mathematical induction, starting with the base case \(n = 6\) where \(256 < 5040\). The inductive hypothesis assumes \(2^{k+2} < (k+1)!\) and demonstrates that \(2^{k+3} < (k+2)!\) holds true by leveraging the factorial growth rate. The conclusion is that the left-hand side grows at a slower rate than the right-hand side as \(n\) increases.
PREREQUISITESMathematicians, educators, students in advanced mathematics courses, and anyone interested in proofs involving inequalities and induction methods.
@ciencero, at this site, use double $ characters at each end for standalone LaTeX, or double # characters at each end for inline LaTeX.ciencero said:$2^{n+2} < (n+1)!$ for all n $\geq 6$
Step 1: For n = 6,
$256 < 5040$.
We assume
$2^{k+2} < (k+1)!$
Induction step:
$2 * 2^{k+2} < 2*(k+1)!$
By noting $2*(k+1)! < (k+2)!$
Then $2^{k+3} < (k+2)!$