Prove by induction

**captainjapan** · November 26th 2008, 04:59 AM

3^n < n! for all n > 6 with n ∈ N.

mathematical induction to prove these?

**Soroban** · November 26th 2008, 06:06 AM

Hello, captainjapan!

Prove by induction: . $3^n \:<\: n! \;\text{ for all }n > 6,\;\;n \in N$

Verify the base case . . .
. . $S(7)\!:\;\;3^7 \:<\:7! \quad\Rightarrow\quad 2187 \:<\:5040$ . . . True

Assume $S(k)\!:\;\;3^k \:<\:k!$

Multiply both sides by 3: . $3^k\cdot3 \:<\:3\cdot k! \quad\Rightarrow\quad 3^{k+1} \:<\:3\cdot k!$ .[1]

. . Since $k > 6\!:\;\;3 \:<\:k$

. . Multiply by $k!\!:\;\;3\cdot k! \:<\:k\cdot k! \quad\Rightarrow\quad 3\cdot k! \:<\:(k+1)!$ .[2]

Combine [1] and [2]: . $3^{k+1} \:<\:3\cdot k! \:<\:(k+1)!$

. . Therefore: . $3^{k+1}\:<\:(k+1)!$

And we have proved $S(k+1)$ . . . The inductive proof is complete.

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