Show $(2n+2)!\geq(n+2)(n+2)!$, $\forall n\in\mathbb{N}$.

Question

It's the last step in a proof, and I just need to show that $$(2n+2)!\geq(n+2)(n+2)!$$ $\forall n\in\mathbb{N}$. I can't seem to do it though, any thoughts?

Answer 1

Just use the monotonicity of the factorial: $$(n+2)(n+2)! \le (n+3)(n+2)! = (n+3)! \le (2n+2)!$$ Where the last step uses $n+3\le 2n+2 \quad\forall\ n\in\mathbb N$

Answer 2

Hint:

$$(2n+2)! \geq (n+3)! \geq (n+2) (n+2)!$$

Answer 3

Since $n \ge 1$, we have $n + n \ge 1 + n \iff 2n - 1 \ge n \iff 2n + 2 \ge n + 3$.
So, $(2n + 2)! \ge (n + 3)! = (n + 3)(n + 2)! \ge (n+2)(n+2)!$ and by transitivity of $\ge$ we conclude that $(2n + 2)! \ge (n+2)(n+2)!$.

Answer 4

Two ways to do it: directly, or by induction.

Direct Proof

Notice that $(n+2)(n+2)!\le(n+3)(n+2)!=(n+3)!$ (by definition).
And clearly: $(n+3)!\le(2n+2)!$
Therefore: $(2n+2)!\ge(n+2)(n+2)!$

---

(Last step of a) proof by induction

Assume that $(2n+2)!\ge(n+2)(n+2)!$, and consider the step $n+1$.

On the one hand, you have: $(2(n+1)+2)!=(2n+4)!=(2n+4)(2n+3)(2n+2)!$
By induction hypothesis: $(2n+4)!\ge(2n+4)(2n+3)(n+2)(n+2)!$

On the other hand, you have: $(n+1+2)(n+1+2)!=(n+3)(n+3)!=(n+3)(n+3)(n+2)!$

By comparison, it is very clear that: $(n+3)(n+3)(n+2)!\le(2n+4)(2n+3)(n+2)(n+2)!$

Such that finally: $(2n+4)!\ge(n+3)(n+3)!$ , which is the property for the step $n+1$.

---

Of course, the first proof is much more elegant.