The discussion revolves around proving the inequality \(2^{n+2} < (n+1)!\) for all \(n \geq 6\) using mathematical induction. Participants are examining the validity of the induction steps and the behavior of both sides of the inequality as \(n\) increases.
Some participants express confidence in the approach, while others point out that \(n = 6\) is not the first instance where the inequality holds. The discussion is ongoing, with various interpretations being explored regarding the growth of the factorial compared to the exponential function.
There is a note about the formatting of LaTeX in the forum, indicating that participants are encouraged to use specific symbols for clarity in mathematical expressions.
@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 $2^{k+3} < (k+2)!$