Let P(n) : 3^n

Question

(A) We want to find the smallest $\lambda \in N$ such that $3^n < n!$ for all $n \geq \lambda$.Check values of $n$:For $n=1: 3^1 = 3, 1! = 1$. ($3 < 1

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) We want to find the smallest $\lambda \in N$ such that $3^n Check values of $n$:For $n=1: 3^1 = 3, 1! = 1$. ($3 For $n=2: 3^2 = 9, 2! = 2$. ($9 For $n=3: 3^3 = 27, 3! = 6$. ($27 For $n=4: 3^4 = 81, 4! = 24$. ($81 For $n=5: 3^5 = 243, 5! = 120$. ($243 For $n=6: 3^6 = 729, 6! = 720$. ($729 For $n=7: 3^7 = 2187, 7! = 5040$. ($2187 Since the inequality holds for $n=7$ and the growth rate of $n!$ is faster than $3^n$,it will hold for all $n \geq 7$.Thus,the smallest value of $\lambda$ is $7$.

Let $P(n) : 3^n < n!$ for $n \in N$ be true for $n \geq \lambda$. Find the smallest value of $\lambda$.

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