# Prove the convergence of a limit

1. Dec 17, 2012

### Felafel

1. The problem statement, all variables and given/known data
it should be all right this time, but could you please check my solution?

prove the convergence and find the limit of the following sequence:
$a_1>0$
$a_{n+1}= 6 \frac{1+a_n}{7+a_n}$
with $n \in \mathbb{N}^*$

3. The attempt at a solution
the sequence is increasing, and its two possible limits are the solution of the associated equation: 2; -3.
However, I have to exclude -3, because being $a_1>0$ and the sequence increasing, it can't converge to a negative number.
Saying it converges is equal to saying that:
for every $\epsilon<0$ there exists a n* s.t. for every n≥n* we have:
$|a_n-L|<\epsilon$
being L=2 in this case and $a_{n+1}>a_n>a_{n*}$ we have
$\frac{6+6a_n}{7+a_n}-2|<\epsilon$
and
$|\frac{6- 14 -2a_n+6a_n}{7+a_n}|<\epsilon$
which holds for every
$a_n> \frac{-7\epsilon-8}{4-\epsilon}$

2. Dec 17, 2012

### SammyS

Staff Emeritus
$\displaystyle \quad\quad \quad\quad \left|\frac{6+6a_n}{7+a_n}-2\right|<\epsilon$
You have a few typos in your post. I cleaned some up and enlarged some fractions.

$\displaystyle \frac{-7\epsilon-8}{4-\epsilon}\ \$ is negative for ε < 4 .

If a1 > 2 , then the sequence is decreasing.

3. Dec 18, 2012

### Felafel

great, thanks!
would it be better then, if i wrote:
which holds for
$\displaystyle \frac{-7\epsilon-8}{4-\epsilon}>0$
with
$\epsilon>4$
?

4. Dec 18, 2012

### SammyS

Staff Emeritus
No. Not at all.

You need to show that $\displaystyle \ \left|\frac{6+6a_n}{7+a_n}-2\right|<\epsilon\ \$ for all ε > 0. The crucial situation is when ε is very small.

5. Dec 18, 2012

### Felafel

should i put it like that:
$|\frac{7\epsilon+8}{4-\epsilon}| > a_n$ which for $\epsilon$ going to 0, goes to 2? --> $a_n<2$

6. Dec 18, 2012

### SammyS

Staff Emeritus
What you need to show is that for all n > n*, $\displaystyle |a_n-2|<\varepsilon\ .$

If you can find a way to define an in terms of n and a1, then the ε-n* proof described above is appropriate.

I have worn out WolramAlpha, messing around with this problem.

I haven't worked with recursively defined sequences lately... but here's an idea.

If the sequence an is monotonically increasing, which it is if 0 < a1 < 2, and if the sequence has an upper bound of 2, then all you need to show is that for ε > 0, there is a n>0 for which ak > 2-ε .

Now that I consider that, that's not easy to do either.

7. Dec 19, 2012

### pasmith

Acutally if $a_1 > 2$ then the sequence is monotonic decreasing. What is true is that if $a_n > 0$ then $a_{n+1} > 0$, so if $a_1 > 0$ then certainly the sequence cannot converge to -3.

With sequences defined by recurrence relations, it's best to note that if $|a_n - L|$ is monotonic decreasing, then it is bounded below (by 0, as absolute values always are) and so $|a_n - L|$ must converge to a limit. Hence $a_n$ converges, and given how $L$ was determined $a_n$ must converge to $L$.

The condition for $|a_n - 2|$ to be monotonic decreasing is that $|a_{n+1} - 2| < |a_n - 2|$ for all n. Now
$$\left|a_{n+1} - 2\right| = \left|6\frac{a_n+1}{a_n+7} - 2\right| = \left|\frac{4a_n - 8}{a_n + 7}\right| = \frac{4}{|a_n + 7|}|a_n - 2|$$
which means that $|a_{n+1} - 2| < |a_n - 2|$ if $|a_n + 7| > 4$. This holds, in particular, if $a_n > 0$. Since $a_1 > 0$ and if $a_n > 0$ then $a_{n+1} > 0$, $|a_{n+1} - 2| < |a_n - 2|$ for all n. Hence $a_n$ tends to a limit, which in the circumstances must be 2.