Induction. Prove: $n! \ge 3^{n-1}/2$ for $n\geq1$

Question

Induction. Prove: $n! \ge 3^{n-1}/2$ for $n\geq1$

For $n = 1$, $1! \geq 1/2$ which is true.

When I start with induction I can't prove it for all natural numbers. I really appreciate your help.

Answer 1

You want to show that $$ n! \ge 3^{(n-1)}/2$$

The problem begs for Mathematical Induction proof.

The statement it true for $n=1$ and $n=2$

If the statement is true for $k\ge 2$ , then

$$(k+1)!= (k+1)k! \ge (k+1)( 3^{(k-1)}/2)\ge 3( 3^{(k-1)}/2) = 3^{k}/2$$

Thus the statement is true for all $n\ge 1.$

Answer 2

Alt. hint:   for $\,n \ge 2\,$:

$$ \dfrac{(n+1)!}{n!} = n+1 \ge 3 = \dfrac{3^n}{3^{n-1}} \;\;\implies\;\;\dfrac{(n+1)!}{3^n} \ge \dfrac{n!}{3^{n-1}} \ge \dfrac{(n-1)!}{3^{n-2}} \ge \ldots \ge \dfrac{2!}{3^1} \gt \dfrac{1}{2} $$

Answer 3

Induction step:

Assume $k!\ge 3^{\frac {k-1}2}$. Can you prove:

$(k+1)! = k!(k+1) \ge 3^{\frac {k-1} 2}(k+1) \underbrace{\ge}_{\text{prove this}} 3^{\frac {k} 2}$?

Note: $3^{\frac k2} = 3^{\frac {k-1}{2}}3^{\frac 12}$.

So can you prove that if $k \ge 1$ that $k + 1 \ge \sqrt{3}$?

Can you prove if $k \ge 1$ then $(k+1)^2 \ge 3$?

Answer 4

Problem statement: Prove using induction that $$n! \geq \dfrac{3^{n-1}}{2}; \forall n \geq 1$$

Solution

• Base case: For $n = 1$

$$1! = 1 \geq \dfrac{3^{1-1}}{2} = \dfrac{1}{2} \tag{1}$$ which is true.

• Assume that the given relation is true of any $n \geq 1$.
• The for $n + 1$, we have \begin{align} (n + 1)! = (n + 1) \times n! \geq \dfrac{3^{n+1-1}}{2} = \dfrac{3^n}{2} = \dfrac{3 \times 3^{n- 1}}{2} = 3 \times \dfrac{3^{n-1}}{2} \end{align} The relation ($\geq$) above is true as using the relation, $n! \geq \dfrac{3^{n-1}}{2}$ and also using the base case, the minimum value possible for $n$ after the base case is $n = 2$, and $n + 1 = 2 + 1 \geq 3$. Hence this proves the claim.