A tough Sequence

  • Thread starter Shing
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  • #1
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Main Question or Discussion Point

[tex]a_{n+1}=3-\frac{1}{a_n}[/tex]
[tex]a_1=1[/tex]

How should I prove that [tex]a_{n+1}>a_n[/tex]
for all n, ?
I have tired to use counter example,
assuming
[tex]a_n>a_{n+1}[/tex]
then contradiction appears,
but I found that what I was doing is just to counter "for all n"



How should I prove that
[tex]a_{n+1}>a_n[/tex] ,for all n ?
 

Answers and Replies

  • #2
52
0
Induction seems to be the obvious choice. Try working backwards by assuming:
[tex]a_{n+1} > a_n[/tex]
and then showing that it implies [tex]a_n > a_{n-1}[/tex]. You should then be able to reverse the steps, prove [tex]a_2 > a_1[/tex] and the induction is complete.
 
  • #3
144
1
probably I got it now,
thanks!
 
Last edited:
  • #4
rock.freak667
Homework Helper
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Well if I had to do this in an exam, I wouldn't use induction,too much for me to write and I don't like to write.

[tex]
a_{n+1}=3-\frac{1}{a_n}[/tex]

[itex]-a_n[/itex] i.e. minus a_n from both sides of the equation.

[tex]
a_{n+1}-a_n =3-\frac{1}{a_n} -a_n[/tex]

Bring the right side to the same denominator and see what happens.From the expression for [itex]a_{n+1}[/itex] and [itex]a_1=1[/itex], what does that say for [itex]a_n[/itex]?
 
  • #5
144
1
Well if I had to do this in an exam, I wouldn't use induction,too much for me to write and I don't like to write.

[tex]
a_{n+1}=3-\frac{1}{a_n}[/tex]

[itex]-a_n[/itex] i.e. minus a_n from both sides of the equation.

[tex]
a_{n+1}-a_n =3-\frac{1}{a_n} -a_n[/tex]

Bring the right side to the same denominator and see what happens.From the expression for [itex]a_{n+1}[/itex] and [itex]a_1=1[/itex], what does that say for [itex]a_n[/itex]?
If you mean [itex]a_{n+1}-a_n[/itex] is positive as [itex]\frac{{a_n}^2+3a_n-1}{a_n}[/itex] is positive and that implies [itex]a_{n+1}>a_n[/itex]

but how do you know that [itex]\frac{{a_n}^2+3a_n-1}{a_n}[/itex] is positive?
or [itex]{a_n}^2+3a_n>1[/itex]?
 
  • #6
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If you want to avoid typing out an inductive argument, simply assume the minimality of n.

Assume for contradiction that there exist at least one n such that [itex]a_{n} \leq a_{n+1}[/itex]. Let [itex]m[/itex] be the smallest integer such integer. Then:
[tex]a_{m} \leq a_{m+1} \Rightarrow 3-\frac{1}{a_{m-1}} \leq 3-\frac{1}{a_m} \Rightarrow a_{m-1} \leq a_m[/tex]
This contradicts the minimality of m.
 
  • #7
144
1
If you want to avoid typing out an inductive argument, simply assume the minimality of n.

Assume for contradiction that there exist at least one n such that [itex]a_{n} \leq a_{n+1}[/itex]. Let [itex]m[/itex] be the smallest integer such integer. Then:
[tex]a_{m} \leq a_{m+1} \Rightarrow 3-\frac{1}{a_{m-1}} \leq 3-\frac{1}{a_m} \Rightarrow a_{m-1} \leq a_m[/tex]
This contradicts the minimality of m.
What is the minimality of n?
I checked this in the Internet, is it something about Strong minimality theory??
 
  • #8
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No it's simply a way to say that m is the least possible. We assumed that [itex]m[/itex] was the smallest value for which the original statement was false, but we showed that this implies that [itex]m-1[/itex] makes the original statement false. There exist a value smaller than m even though we assumed m was the smallest, therefore there can't exist a smallest element.
 
  • #9
144
0
edit: Sorry, was wrong :)
 
  • #10
73
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Direct Proof?

[tex]a_{n+1}=3-\frac{1}{a_n}[/tex]
[tex]a_1=1[/tex]

Prove: [itex]a_{n+1}>a_n[/itex] for all n?

Let's use a direct comparison:

[tex] a_{n+1} > a_n [/tex]
[tex] 3 - \frac{1}{a_n} > a_n [/tex]
[tex] \frac{a_n^2-3a_n+1}{a_n} < 0 [/tex]

This is zero or undefined for [itex] a_n = \frac{3 \pm \sqrt{5}}{2}, 0 [/itex]

The solution to this inequality is [itex] a_n < \frac{3 - \sqrt{5}}{2} \cup 0 < a_n < \frac{3 + \sqrt{5}}{2}[/itex].

Since [itex]a_1[/itex] is in the solution set, this is shown to be true.

Now we have to show that you can't jump boundaries for [itex] a_1 = 1 [/itex].
 
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  • #11
73
0
I made a mistake in my initial reply here it is fixed.

[tex]a_{n+1}=3-\frac{1}{a_n}[/tex]
[tex]a_1=1[/tex]

Prove: [itex]a_{n+1}>a_n[/itex] for all n?

Let's use a direct comparison:

[tex] a_{n+1} > a_n [/tex]
[tex] 3 - \frac{1}{a_n} > a_n [/tex]
[tex] \frac{a_n^2-3a_n+1}{a_n} < 0 [/tex]

This is zero or undefined for [itex] a_n = \frac{3 \pm \sqrt{5}}{2}, 0 [/itex]

The solution to this inequality is [itex] a_n < 0 \cup \frac{3 - \sqrt{5}}{2} < a_n < \frac{3 + \sqrt{5}}{2}[/itex].

Since [itex]a_1[/itex] is in the solution set, this is shown to be true.

Now we have to show that you can't jump boundaries for [itex] a_1 = 1 [/itex].
 
  • #12
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Since [itex]a_1 = 1[/itex] lies in the second boundary, we will only work with that boundary.

Next we have to find all values for [itex] a_n [/itex] that will allow [itex] \frac{3-\sqrt{5}}{2} < 3 - \frac{1}{a_n} < \frac{3+\sqrt{5}}{2} [/itex] to be true.

[tex]\frac{3-\sqrt{5}}{2} < 3 - \frac{1}{a_n}[/tex]
[tex] a_n < 0 \cup a_n > \frac{3-\sqrt{5}}{2}[/tex]

[tex] \frac{3+\sqrt{5}}{2} > 3 - \frac{1}{a_n}[/tex]
[tex] 0 < a_n < \frac{3+\sqrt{5}}{2} [/tex]

The intersection of those intervals yields: [itex] \frac{3-\sqrt{5}}{2} < a_n < \frac{3+\sqrt{5}}{2}[/itex]

Since [itex]a_1=1[/itex] is in our interval we know that [itex]a_{n+1}[/itex] will not jump out of the good interval.

Is this good enough?
 
Last edited:

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