# A tough Sequence

## Main Question or Discussion Point

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

How should I prove that $$a_{n+1}>a_n$$
for all n, ?
I have tired to use counter example,
assuming
$$a_n>a_{n+1}$$
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 ?

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.

probably I got it now,
thanks!

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rock.freak667
Homework Helper
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.

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

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

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

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 $a_n$?

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.

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

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

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

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 $a_{n+1}>a_n$

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

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 $a_{n} \leq a_{n+1}$. 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.

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??

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.

edit: Sorry, was wrong :)

Direct Proof?

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

Prove: $a_{n+1}>a_n$ for all n?

Let's use a direct comparison:

$$a_{n+1} > a_n$$
$$3 - \frac{1}{a_n} > a_n$$
$$\frac{a_n^2-3a_n+1}{a_n} < 0$$

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

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

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 $a_1 = 1$.

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

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

Prove: $a_{n+1}>a_n$ for all n?

Let's use a direct comparison:

$$a_{n+1} > a_n$$
$$3 - \frac{1}{a_n} > a_n$$
$$\frac{a_n^2-3a_n+1}{a_n} < 0$$

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

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

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 $a_1 = 1$.

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 $\frac{3-\sqrt{5}}{2} < 3 - \frac{1}{a_n} < \frac{3+\sqrt{5}}{2}$ to be true.

$$\frac{3-\sqrt{5}}{2} < 3 - \frac{1}{a_n}$$
$$a_n < 0 \cup a_n > \frac{3-\sqrt{5}}{2}$$

$$\frac{3+\sqrt{5}}{2} > 3 - \frac{1}{a_n}$$
$$0 < a_n < \frac{3+\sqrt{5}}{2}$$

The intersection of those intervals yields: $\frac{3-\sqrt{5}}{2} < a_n < \frac{3+\sqrt{5}}{2}$

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?

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