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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?

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

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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.

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  • $\begingroup$ 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 $\endgroup$ – pic Feb 25 at 8:54
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    $\begingroup$ 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. $\endgroup$ – fleablood Feb 25 at 9:04
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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

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

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  • $\begingroup$ We silently used the property 1≤n!. you use this on green hightlight right? $\endgroup$ – pic Feb 25 at 8:38
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    $\begingroup$ @pic: not at all, this is from inductive hypothesis. $\endgroup$ – Yves Daoust Feb 25 at 8:38
  • $\begingroup$ 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 ? $\endgroup$ – pic Feb 25 at 8:59
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    $\begingroup$ @pic: I have completely changed my answer. $\endgroup$ – Yves Daoust Feb 25 at 9:29

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