# This is sufficient to prove by induction that 2n < n! for n >= 4?

I start with base saying that this is valid for one by simple algebraic calculus:

This is my base for $$n = 4$$ $$2 \cdot 4 \leq 4! \Leftrightarrow 8 \leq 24$$

In my hypotheses just say that exists a number $$k$$ that satisfies the expression for some $$k\geq 4$$: $$2k < k!$$

And my inductive step I multiply both sides from hypotheses by $$(k+1)$$ as follows: $$(k+1) \cdot 2k < k! \cdot (k+1)$$ $$\Leftrightarrow (k+1) \cdot 2k < (k+1)!$$

So by hypotheses we know that $$k\geq 4$$ soon can afirms that $$2k(k+1) > 2k$$. Given that $$(k+1)! > 2k(k+1)$$ and $$2k(k+1) > 2k$$ is proved by induction that $$2n < n!$$

I wrong anything or is insufficient to prove the enunciated?

• In the inductive step you have toshow that $2(k+1) < (k+1)!$ – Fred Feb 25 at 8:23
• You need to add an additional step $(k+1)\cdot2k>(k+1)\cdot2$ which is pretty obvious. – user Feb 25 at 8:24
• Seems hard to build an inductive proof that avoids the simplification $2n<n!\Leftarrow 2<(n-1)!$, a non-inductive proof. – Yves Daoust Feb 25 at 8:25
• "soon can afirms that 2k(k+1)>2k" You don't want to affire $2k(k+1) > 2k$. You want to affirm $2k(k+1) > 2(k+1)$. – fleablood Feb 25 at 8:39

## 3 Answers

It's often better to prove it in one line:

$$2(k+1) = 2k + 2 < k!+2 < k!\cdot (k+1) = (k+1)!$$,

where the first inequality is by induction hypothesis and the second inequality holds for $$k\geq 4$$ anyway (for completeness, it could also be proved by induction).

Fixe that $$2k(k+1) > 2(k+1)$$ and not $$2k(k+1) > 2k$$ and your proof is fine.

Alternatively it might be simpler to go left to right rather than right to left

$$2k < k!$$

$$2k + 2 < k! + 2 < k! + 2\cdot 3 < k! + 2\cdot 3\cdot 4 < ...... < k! + k! =$$

$$2k! < (k+1)k! = (k+1)!$$.

....

You know.... as far as inequalities go... this aint even close.

• Yes this was my second attempt, first I tried get out from $2(k+1) < (k+1)!$ to $2k < k!$ but I spended a lot time and move to on with this attempt from right to left – pic Feb 25 at 8:54
• It's always better to go from what you know ($2k < k!$) to what you want to prove ($2(k+1) < (k+1)!$) rather than the other way around. – fleablood Feb 25 at 9:04

A simpler proof:

For all naturals, $$2<(n-1)!\implies 2n< n!$$ and it suffices to prove

$$n\ge3\implies n!>2.$$

Very easily,

$$3!>2$$ and

$$n!>2\implies (n+1)n!>2.$$

• We silently used the property 1≤n!. you use this on green hightlight right? – pic Feb 25 at 8:38
• @pic: not at all, this is from inductive hypothesis. – Yves Daoust Feb 25 at 8:38
• Ok, now I see, but the last inequality shouldn't be a $n!+n \leq n!+n!n$ given that $n!$ on $n!n$ can be one ? – pic Feb 25 at 8:59
• @pic: I have completely changed my answer. – Yves Daoust Feb 25 at 9:29