3^n < n! for all n > 6 with n ∈ N.
mathematical induction to prove these?
Hello, captainjapan!
Verify the base case . . .Prove by induction: .$\displaystyle 3^n \:<\: n! \;\text{ for all }n > 6,\;\;n \in N$
. . $\displaystyle S(7)\!:\;\;3^7 \:<\:7! \quad\Rightarrow\quad 2187 \:<\:5040$ . . . True
Assume $\displaystyle S(k)\!:\;\;3^k \:<\:k!$
Multiply both sides by 3: .$\displaystyle 3^k\cdot3 \:<\:3\cdot k! \quad\Rightarrow\quad 3^{k+1} \:<\:3\cdot k! $ .[1]
. . Since $\displaystyle k > 6\!:\;\;3 \:<\:k$
. . Multiply by $\displaystyle k!\!:\;\;3\cdot k! \:<\:k\cdot k! \quad\Rightarrow\quad 3\cdot k! \:<\:(k+1)!$ .[2]
Combine [1] and [2]: .$\displaystyle 3^{k+1} \:<\:3\cdot k! \:<\:(k+1)!$
. . Therefore: .$\displaystyle 3^{k+1}\:<\:(k+1)!$
And we have proved $\displaystyle S(k+1)$ . . . The inductive proof is complete.