Prove the convergence of a limit

[Felafel]

Homework Statement

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 $n \in \mathbb{N}^*$

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
$a_n> \frac{-7\epsilon-8}{4-\epsilon}$

 

[SammyS]

Homework Statement

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:
$a_1>0$
$\displaystyle a_{n+1}= 6 \frac{1+a_n}{7+a_n}\ \$ with $n \in \mathbb{N}^*$

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
$\displaystyle \frac{6+6a_n}{7+a_n}-2|<\epsilon$

$\displaystyle \quad\quad \quad\quad \left|\frac{6+6a_n}{7+a_n}-2\right|<\epsilon$

and
$\displaystyle \left|\frac{6- 14 -2a_n+6a_n}{7+a_n}\right|<\epsilon$
which holds for every
$a_n> \frac{-7\epsilon-8}{4-\epsilon}$

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.

 

[Felafel]

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

If a₁ > 2 , then the sequence is decreasing.

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

 

[SammyS]

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

No. Not at all.

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

 

[Felafel]

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

 

[SammyS]

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

What you need to show is that for all n > n^*, \displaystyle |a_n-2|&lt;\varepsilon\ .

If you can find a way to define a_n 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 a_n 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 a_k > 2-ε .

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

 

[pasmith]

Homework Statement

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 $n \in \mathbb{N}^*$

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.

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

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

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