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Prove the statement is true using mathematical induction: $$2n-1 \leq n!$$

My attempt: this is true for $n=1$.

Suppose it is true for $n$, i.e., $2n-1 \leq n!$

Now, $2n-1 \leq n!\implies 2n-1+2 \leq n!+2$

From here, how do I proceed?

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  • $\begingroup$ Hint: $(n + 1)! = (n + 1) \cdot n!$ $\endgroup$ – iamwhoiam Sep 7 '17 at 3:18
  • $\begingroup$ $(n+1)!-n!=n.n!>2n!>2$ $\endgroup$ – Nosrati Sep 7 '17 at 3:19
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$$\forall n\geq2, \ (n+1)!=n!(n+1)\geq(2n-1)(n+1)\geq2n+1$$

but for $n=2$ our statement is false, which says that $ \ \forall n\geq 3, \ n!\geq2n-1$.

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  • $\begingroup$ And $2n + 1 > 2n - 1$ $\endgroup$ – Mr Pie Sep 7 '17 at 3:22
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for $n+1$ we get $2(n+1)-1=2n+1=2n-1+3 \le n!+3\le n!(n+1)$, which is true for $n \ge 2$

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  • $\begingroup$ The relation given by OP is false for n=2...it starts to be true at n=3 $\endgroup$ – Aryadeva Sep 7 '17 at 3:40

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