Prove 2^n < (n+2)!; \forall n \ge 0 by Induction

  • Thread starter jonroberts74
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In summary, to prove 2^n < (n+2)! for all n greater than or equal to 0, we use induction. We first show that the statement is true for the base case n = 0, then assume it is true for some k and use that to prove it for k+1. We also use the fact that 2 < k+3 as a key step in the proof. This leads us to the conclusion that 2^n < (n+2)! for all n greater than or equal to 0.
  • #1
jonroberts74
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Homework Statement

prove by induction

[tex] 2^n < (n+2)!; \forall n \ge 0[/tex]

P(0)

[tex] 2^0 < (0+2)! [/tex] [easy]

P(k)

[tex]2^k<(k+2)![/tex]
[tex]= 2^k < (k+2)(k+1)k![/tex]

P(k+1)

[tex]2^{k+1} < (k+1+2)! [/tex]
[tex]= 2^k \cdot 2 < (k+3)(k+2)(k+1)k![/tex]

its pretty clear that

[tex]2 < k+3[/tex]

how do I show that though
 
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  • #2
jonroberts74 said:
its pretty clear that

[tex]2 < k+3[/tex]

how do I show that though

I would personally just say it's obvious. But if you want, you can prove it by induction. It all depends on how you defined ##<## and what your axioms are.
 
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  • #3
I see you recognize that you cannot just write "[itex]2^{k+1}<((k+1)+ 2)![/itex]" since that is what you want to find- you wrote it only to remind yourself what you want to arrive at.

Your "induction hypothesis" is that [itex]2^k< (k+ 2)![/itex]. Now write [itex]2^{k+1}= 2(2^k)< 2(k+2)![/itex].

Now, use your "2< k+ 3" to get [itex]2^{k+1}< 2(k+2)!< (k+2)!(k+ 3)= (k+3)!= (k+1+2)![/itex].

To prove that 2< k+ 3, start from the fact that, since [itex]0\le k[/itex], [itex]-1< k[/itex] and add 3 to both sides.
 
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  • #4
HallsofIvy said:
I see you recognize that you cannot just write "[itex]2^{k+1}<((k+1)+ 2)![/itex]" since that is what you want to find- you wrote it only to remind yourself what you want to arrive at.

Your "induction hypothesis" is that [itex]2^k< (k+ 2)![/itex]. Now write [itex]2^{k+1}= 2(2^k)< 2(k+2)![/itex].

Now, use your "2< k+ 3" to get [tex]2^{k+1}< 2(k+2)!< (k+2)!(k+ 3)= (k+3)!= (k+1+2)![/tex].

To prove that 2< k+ 3, start from the fact that, since [itex]0\le k[/itex], [itex]-1< k[/itex] and add 3 to both sides.


I am taking this

[itex]2^{k+1}= 2(2^k)< 2(k+2)![/itex]

by hypothesis, adding the multiple 2 to each side doesn't change the inequality

then prove 2<k+3

and do this from the fact

[tex]0 \le k[/tex] so -1<k ---> -1+3<k+3 = 2< K+3

then I can say [itex] 2(2^k)< 2(k+2)! < (k+2)!(k+3)[/itex].
 

1. How does induction work in proving this statement?

Induction is a mathematical proof technique that involves proving a statement for a base case and then showing that if the statement holds for a particular value, it also holds for the next value. In this case, we will prove the statement for n = 0 and then show that if it holds for n, it also holds for n+1.

2. Why is it important to prove this statement?

Proving this statement is important because it shows that for all values of n greater than or equal to 0, the expression 2^n < (n+2)! is true. This can be useful in various mathematical and scientific applications.

3. What is the significance of the inequality in this statement?

The inequality 2^n < (n+2)! means that the value of 2^n is always less than the value of (n+2)!, regardless of the value of n. This relationship is important in understanding the growth rate of these two expressions.

4. How can we prove this statement using mathematical induction?

To prove this statement using mathematical induction, we will first show that the statement is true for the base case n=0. Then, we will assume that the statement is true for some arbitrary value of n and use this assumption to prove that it is also true for n+1. This will complete the proof by showing that the statement holds for all values of n greater than or equal to 0.

5. Can this statement be proven using other techniques besides induction?

Yes, this statement can be proven using other techniques such as direct proof, contradiction, or contrapositive. However, mathematical induction is the most commonly used technique for proving statements about natural numbers and can often be the most efficient method for proving this type of statement.

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