How Can I Prove that a_{n+1}>a_n for All n in a Tough Sequence?

[Shing]

a_{n+1}=3-\frac{1}{a_n}
a_1=1

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



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

 

[gunch]

Induction seems to be the obvious choice. Try working backwards by assuming:
a_{n+1} > a_n
and then showing that it implies a_n > a_{n-1}. You should then be able to reverse the steps, prove a_2 > a_1 and the induction is complete.

 

[Shing]

probably I got it now,
thanks!

 

[rock.freak667]

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.

<br /> a_{n+1}=3-\frac{1}{a_n}

-a_n i.e. minus a_n from both sides of the equation.

<br /> a_{n+1}-a_n =3-\frac{1}{a_n} -a_n

Bring the right side to the same denominator and see what happens.From the expression for a_{n+1} and a_1=1, what does that say for a_n?

 

[Shing]

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.

<br /> a_{n+1}=3-\frac{1}{a_n}

-a_n i.e. minus a_n from both sides of the equation.

<br /> a_{n+1}-a_n =3-\frac{1}{a_n} -a_n

Bring the right side to the same denominator and see what happens.From the expression for a_{n+1} and a_1=1, what does that say for a_n?

If you mean a_{n+1}-a_n is positive as \frac{{a_n}^2+3a_n-1}{a_n} is positive and that implies a_{n+1}&gt;a_n

but how do you know that \frac{{a_n}^2+3a_n-1}{a_n} is positive?
or {a_n}^2+3a_n&gt;1?

 

[gunch]

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 a_{n} \leq a_{n+1}. Let m be the smallest integer such integer. Then:
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
This contradicts the minimality of m.

 

[Shing]

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 a_{n} \leq a_{n+1}. Let m be the smallest integer such integer. Then:
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
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??

 

[gunch]

No it's simply a way to say that m is the least possible. We assumed that m was the smallest value for which the original statement was false, but we showed that this implies that m-1 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.

 

[Kurret]

edit: Sorry, was wrong :)

 

[sennyk]

Direct Proof?

a_{n+1}=3-\frac{1}{a_n}
a_1=1

Prove: a_{n+1}&gt;a_n for all n?

Let's use a direct comparison:

a_{n+1} &gt; a_n
3 - \frac{1}{a_n} &gt; a_n
\frac{a_n^2-3a_n+1}{a_n} &lt; 0

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

The solution to this inequality is a_n &lt; \frac{3 - \sqrt{5}}{2} \cup 0 &lt; a_n &lt; \frac{3 + \sqrt{5}}{2}.

Since a_1 is in the solution set, this is shown to be true.

Now we have to show that you can't jump boundaries for a_1 = 1.

 

[sennyk]

I made a mistake in my initial reply here it is fixed.

a_{n+1}=3-\frac{1}{a_n}
a_1=1

Prove: a_{n+1}&gt;a_n for all n?

Let's use a direct comparison:

a_{n+1} &gt; a_n
3 - \frac{1}{a_n} &gt; a_n
\frac{a_n^2-3a_n+1}{a_n} &lt; 0

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

The solution to this inequality is a_n &lt; 0 \cup \frac{3 - \sqrt{5}}{2} &lt; a_n &lt; \frac{3 + \sqrt{5}}{2}.

Since a_1 is in the solution set, this is shown to be true.

Now we have to show that you can't jump boundaries for a_1 = 1.

 

[sennyk]

Since a_1 = 1 lies in the second boundary, we will only work with that boundary.

Next we have to find all values for a_n that will allow \frac{3-\sqrt{5}}{2} &lt; 3 - \frac{1}{a_n} &lt; \frac{3+\sqrt{5}}{2} to be true.

\frac{3-\sqrt{5}}{2} &lt; 3 - \frac{1}{a_n}
a_n &lt; 0 \cup a_n &gt; \frac{3-\sqrt{5}}{2}

\frac{3+\sqrt{5}}{2} &gt; 3 - \frac{1}{a_n}
0 &lt; a_n &lt; \frac{3+\sqrt{5}}{2}

The intersection of those intervals yields: \frac{3-\sqrt{5}}{2} &lt; a_n &lt; \frac{3+\sqrt{5}}{2}

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

Is this good enough?