##\lim_{n \to \infty} ## for the sequence at ##a_n##

[Helly123]

Homework Statement

$a_1$ = 3
$a_{n+1}$ = $\frac{2}{3} a_n + \frac{1}{4}$

Homework Equations

[/B]

The Attempt at a Solution

Sequence i got :
$a_1, a_2, a_3, a_4, a_5$
$3, \frac{9}{4}, \frac{7}{4}, \frac{17}{12}, \frac{43}{36}$

I tried to find the formula of $a_n$
$a_n = \frac{3}{2} a_{n+1} - \frac{3}{8}$

Can i get a clue? Is it necessary to know if it is geometric or arithmetic sequence?

 

[Dick]

Homework Statement

$a_1$ = 3
$a_{n+1}$ = $\frac{2}{3} a_n + \frac{1}{4}$

Homework Equations

[/B]

The Attempt at a Solution

Sequence i got :
$a_1, a_2, a_3, a_4, a_5$
$3, \frac{9}{4}, \frac{7}{4}, \frac{17}{12}, \frac{43}{36}$

I tried to find the formula of $a_n$
$a_n = \frac{3}{2} a_{n+1} - \frac{3}{8}$

Can i get a clue? Is it necessary to know if it is geometric or arithmetic sequence?

It's neither arithmetic nor geometric. But you don't need to know that to find the limit. IF $a_n$ approaches a limit $L$, then what do you get if you take the limit of $a_{n+1}$ = $\frac{2}{3} a_n + \frac{1}{4}$? What must $L$ be? But that's ASSUMING that there is a limit. Proving there IS a limit takes a different kind of thinking. How might you show that?

 

[fresh_42]

Homework Statement

$a_1$ = 3
$a_{n+1}$ = $\frac{2}{3} a_n + \frac{1}{4}$

Homework Equations

[/B]

The Attempt at a Solution

Sequence i got :
$a_1, a_2, a_3, a_4, a_5$
$3, \frac{9}{4}, \frac{7}{4}, \frac{17}{12}, \frac{43}{36}$

I tried to find the formula of $a_n$
$a_n = \frac{3}{2} a_{n+1} - \frac{3}{8}$

Can i get a clue? Is it necessary to know if it is geometric or arithmetic sequence?

Do you know what makes the difference between the two?
What did you get for $a_{n+1}=a_{n+1}(a_{n-1})=a_{n+1}(a_{n-2})= \ldots ?$

 

[Mark44]

I tried to find the formula of $a_n$
$a_n = \frac{3}{2} a_{n+1} - \frac{3}{8}$

This isn't what is meant by finding the formula for $a_n$. In your formula, to get $a_n$ you need to know $a_{n+1}$. "Finding the formula" means being able to determine $a_n$ in terms of $a_1$.

Do you know what makes the difference between the two? What did you get for $a_{n+1}=a_{n+1}(a_{n-1})=a_{n+1}(a_{n-2})= \ldots ?$

I don't know what to make of this suggestion.

 

[Mark44]

Here's a start:

$a_1 = 3$
$a_2 = \frac 2 3 a_1 + \frac 1 4$
$a_3 = \frac 2 3 (\frac 2 3 a_1 + \frac 1 4) + \frac 1 4$
The part in parentheses is $a_2$, in terms of $a_1$
$a_4 = \frac 2 3 ( \frac 2 3 (\frac 2 3 a_1 + \frac 1 4) + \frac 1 4) + \frac 1 4$
The part in the innermost parentheses is $a_2$, in terms of $a_1$.
I found it helpful to expand (multiply out) the expressions on the right.
Continue with this process, until you can see a pattern for $a_n$.

 

[fresh_42]

I don't know what to make of this suggestion.

The notation $f=f(x)$ reads: $f$ as a function of $x$. So $a_{n+1}=a_{n+1}(a_{n-1})$ means $a_{n+1}$ as a function of $a_{n-1}$. The same what you wrote, just downwards than upwards.

 

[Mark44]

The notation $f=f(x)$ reads: $f$ as a function of $x$. So $a_{n+1}=a_{n+1}(a_{n-1})$ means $a_{n+1}$ as a function of $a_{n-1}$. The same what you wrote, just downwards than upwards.

I get function notation, but I didn't realize you were using function notation on a term of a sequence. I thought you were multiplying $a_{n+1}$ and $a_{n-1}$.

 

[fresh_42]

I get function notation, but I didn't realize you were using function notation on a term of a sequence. I thought you were multiplying $a_{n+1}$ and $a_{n-1}$.

Sorry for laziness. And I hoped a bit the OP would have asked to get the process going.

 

[Helly123]

IF $a_n$ approaches a limit $L$, then what do you get if you take the limit of $a_{n+1}$ = $\frac{2}{3} a_n + \frac{1}{4}$?

$a_n$ approaches a limit $L$. What does it mean? Lim of $a_n$ the same as limit $L$? Or $a_n$ has value of $L$?
And take the limit as x approaches $\infty$ ?

 

[Mark44]

$a_n$ approaches a limit $L$. What does it mean? Lim of $a_n$ the same as limit $L$?

Yes, although the full expression for this is $\lim_{n \to \infty} a_n = L$.

Or $a_n$ has value of $L$?

No, not necessarily. $a_n$ can have L as a limit without any element of the sequence actually equalling L.

And take the limit as x approaches $\infty$ ?

No, as n approaches $\infty$.

 

[Helly123]

Do you know what makes the difference between the two?
What did you get for $a_{n+1}=a_{n+1}(a_{n-1})=a_{n+1}(a_{n-2})= \ldots ?$

Here's a start:

$a_1 = 3$
$a_2 = \frac 2 3 a_1 + \frac 1 4$
$a_3 = \frac 2 3 (\frac 2 3 a_1 + \frac 1 4) + \frac 1 4$
The part in parentheses is $a_2$, in terms of $a_1$
$a_4 = \frac 2 3 ( \frac 2 3 (\frac 2 3 a_1 + \frac 1 4) + \frac 1 4) + \frac 1 4$
The part in the innermost parentheses is $a_2$, in terms of $a_1$.
I found it helpful to expand (multiply out) the expressions on the right.
Continue with this process, until you can see a pattern for $a_n$.

I guess, the formula for $a_n$ will be $(\frac{2}{3})^{(n-1)}a_1 + (n-1)\frac{1}{4}$ = $3.(\frac{2}{3})^{(n-1)} + \frac{1}{4}(n-1)$

Take the limit for $a_n$ as x approaches $\infty$ i don't get what the question wants...

Usually taking the limit as x approaches $\infty$ , i have to substitute the x variable into $\infty$

 

[fresh_42]

And take the limit as x approaches $\infty$ ?

Think of it as $\frac{1}{n} \longrightarrow 0$. This sequence approaches zero, i.e. we can get arbitrary close to zero without ever reaching it.

 

[Helly123]

Sorry. IT HAS TO BE $\lim_{n \to \infty}$ ALL THIS TIME. NOT $\lim_{x \to \infty}$ . i made mistake in the question

 

[fresh_42]

I guess, the formula for $a_n$ will be $(\frac{2}{3})^{(n-1)}a_1 + (n-1)\frac{1}{4}$ = $3.(\frac{2}{3})^{(n-1)} + \frac{1}{4}(n-1)$

To guess is a bad idea. How often will you guess, because this first guess is wrong?

Take the limit for $a_n$ as x approaches $\infty$ i don't get what the question wants...

You have written something with $(\frac{2}{3})^n$. So what's $\lim_{n \to \infty} (\frac{2}{3})^n\,$?

Usually taking the limit as x approaches $\infty$ , i have to substitute the x variable into $\infty$

There is no variable $x$. We only have a counter $n$. Think of the two examples I gave you.
And write down a few more terms of the sequence to see, why $n\cdot \frac{1}{4}$ is wrong.

 

[Helly123]

Yes, although the full expression for this is $\lim_{n \to \infty} a_n = L$.

why on that expression, $\lim_{n \to \infty} a_n = L$ and NOT $\lim_{n \to \infty} a_n = \lim_{n \to \infty} L$

 

[Mark44]

I guess, the formula for $a_n$ will be $(\frac{2}{3})^{(n-1)}a_1 + (n-1)\frac{1}{4}$ = $3.(\frac{2}{3})^{(n-1)} + \frac{1}{4}(n-1)$

No, not even close. If you take the limit of the above, as n grows large, the result approaches infinity. You worked out several terms in the sequence. Does it seem that they are growing largerl without bound. Take a look at my suggestion in post #5 again.

Take the limit for $a_n$ as x approaches $\infty$ i don't get what the question wants...

What you wrote is meaningless, since $a_n$ doesn't have anything to do with x. You want to find out what $a_n$ is doing as n gets large, not x.

Usually taking the limit as x approaches $\infty$ , i have to substitute the x variable into $\infty$

There is no x in this problem.

 

[Mark44]

why on that expression, $\lim_{n \to \infty} a_n = L$ and NOT $\lim_{n \to \infty} a_n = \lim_{n \to \infty} L$

$\lim_{n \to \infty} L$ is just L. Since L doesn't depend on n, it's redundant to write "lim... L". L is just a number.

 

[fresh_42]

why on that expression, $\lim_{n \to \infty} a_n = L$ and NOT $\lim_{n \to \infty} a_n = \lim_{n \to \infty} L$

Because $\lim_{n \to \infty} L = L$. The limit of the sequence $(L,L,L,L,L,L,L,L,L,L,L,L,L,L,...)$ is surprisingly $L$.

 

[Helly123]

Do you know what makes the difference between the two?
What did you get for $a_{n+1}=a_{n+1}(a_{n-1})=a_{n+1}(a_{n-2})= \ldots ?$

I don't understand.. difference between $a_{n+1}(a_{n-1})$ and $a_{n+1}(a_{n-2})$ ?
The difference is changing according to n.
$a_{n+1}(a_{n-1})$ means, i fill a at (n-1) into the formula $a_{n+1}$ right?
Then why the value of $a_{n+1}(a_{n-1})$ = $a_{n+1}(a_{n-2})$ ? Since $a_{n-1}$ not the same as $a_{n-2}$

You have written something with $(\frac{2}{3})^n$. So what's $\lim_{n \to \infty} (\frac{2}{3})^n\,$?

And write down a few more terms of the sequence to see, why $n\cdot \frac{1}{4}$ is wrong.

I tried this
$a_2 - a_1$ =
$-\frac{1}{3}a_{1} + \frac{1}{4}$
Then
$a_3 - a_2$ = $-\frac{1}{3}a_{2} + \frac{1}{4}$
I still have to think again

 

[fresh_42]

I don't understand.. difference between $a_{n+1}(a_{n-1})$ and $a_{n+1}(a_{n-2})$ ?
The difference is changing according to n.
$a_{n+1}(a_{n-1})$ means, i fill a at (n-1) into the formula $a_{n+1}$ right?
Then why the value of $a_{n+1}(a_{n-1})$ = $a_{n+1}(a_{n-2})$ ? Since $a_{n-1}$ not the same as $a_{n-2}$I tried this
$a_2 - a_1$ =
$-\frac{1}{3}a_{1} + \frac{1}{4}$
Then
$a_3 - a_2$ = $-\frac{1}{3}a_{2} + \frac{1}{4}$
I still have to think again

You can as well start from the top. But in any case you will need the law of distribution, because we want to arrive at a function with only $n$ and numbers in it: $a_n = f(n)$

 

[Helly123]

No, not even close. If you take the limit of the above, as n grows large, the result approaches infinity. You worked out several terms in the sequence. Does it seem that they are growing largerl without bound. Take a look at my suggestion in post #5 again.

The sequence getting smaller as n approaches $\infty$ but the formula i guessed is getting the bigger value as n approaches $\infty$ is that what you meant? That is why it's wrong?

 

[Helly123]

You can as well start from the top. But in any case you will need the law of distribution, because we want to arrive at a function with only $n$ and numbers in it: $a_n = f(n)$

Yes. But what do you mean by that 2 examples.. as what i explained in #19. I don't get that

 

[fresh_42]

The sequence getting smaller as n approaches $\infty$ but the formula i guessed is getting the bigger value as n approaches $\infty$ is that what you meant? That is why it's wrong?

Yes, that's the obvious problem with your guess, but I also have calculated the limit and compared the results directly.
I just started with
$$
a_n = \frac{2}{3}a_{n-1}+\frac{1}{4}= \frac{2}{3}\left(\frac{2}{3}a_{n-2}+\frac{1}{4}\right)+\frac{1}{4}=\frac{2}{3}\left(\frac{2}{3}\left(\frac{2}{3}a_{n-3}+\frac{1}{4}\right)+\frac{1}{4}\right)+\frac{1}{4} = \ldots
$$
Of course I calculated this to get rid of the parentheses.

 

[fresh_42]

Yes. But what do you mean by that 2 examples.. as what i explained in #19. I don't get that

That was only to explain what a limit with $n \to \infty$ means, and that we do not have to have the limit itself in the sequence.

 

[Helly123]

Yes, that's the obvious problem with your guess, but I also have calculated the limit and compared the results directly.
I just started with
$$
a_n = \frac{2}{3}a_{n-1}+\frac{1}{4}= \frac{2}{3}\left(\frac{2}{3}a_{n-2}+\frac{1}{4}\right)+\frac{1}{4}=\frac{2}{3}\left(\frac{2}{3}\left(\frac{2}{3}a_{n-3}+\frac{1}{4}\right)+\frac{1}{4}\right)+\frac{1}{4} = \ldots
$$
Of course I calculated this to get rid of the parentheses.

Is there any clue to get the $a_n$ formula with only n and numbers in it?

$$
a_\infty = \frac{2}{3}a_{(\infty-1)}+\frac{1}{4} $$

 

[Mark44]

Is there any clue to get the $a_n$ formula with only n and numbers in it?

$$
a_\infty = \frac{2}{3}a_{(\infty-1)}+\frac{1}{4} $$

You can't do this. For one thing, you can't have an index of $\infty$. You also can't have $\infty$ as part of an arithmetic expression like you have in $\infty - 1$.
fresh_42 and I are approaching this from two different directions: in post #23, he's working from $a_{n+1}$ down (i.e., starting with $a_{n+1}$ in terms of $a_n$, then with $a_{n+1}$ in terms of $a_{n-1}$, then with $a_{n+1}$ in terms of $a_{n-2}$, and so on. In post #5, I working my way from $a_2$ in terms of $a_1$, then to $a_3$ in terms of $a_1$, and then $a_4$ in terms of $a_1$. Do this as many times as necessary until you see a pattern.

 

[fresh_42]

Is there any clue to get the $a_n$ formula with only n and numbers in it?

Yes. Calculate the expressions:
$a_n = \frac{2}{3}a_{n-1}+\frac{1}{4}$ and $a_{n-1} = \frac{2}{3}a_{n-2}+\frac{1}{4}$ result in
$$a_n=\frac{2}{3}\left(\frac{2}{3}a_{n-2}+\frac{1}{4}\right)+\frac{1}{4}=\left(\frac{2}{3}\right)^2 a_{n-2}+\frac{1}{4}\left(\left(\frac{2}{3}\right)^0+ \left(\frac{2}{3}\right)^1 \right)$$
and in the next step with $a_{n-2}= \frac{2}{3}a_{n-3}+\frac{1}{4}$
$$a_n=\frac{2}{3}\left(\frac{2}{3}\left(\frac{2}{3}a_{n-3}+\frac{1}{4}\right)+\frac{1}{4}\right)+\frac{1}{4}=\left(\frac{2}{3}\right)^3 a_{n-3}+\frac{1}{4}\left(\left(\frac{2}{3}\right)^0+ \left(\frac{2}{3}\right)^1 + \left(\frac{2}{3}\right)^2 \right)$$
Now look at the pattern, go on until $a_n=\ldots \, \cdot a_1 + \ldots$, substitute $a_1=3$ and compute $\sum_{i=0}^n q^n$ with $q=\frac{2}{3}$. This sum can be calculated by $(x-1)^n=(x-1)\cdot (x^{n-1}+x^{n-2}+\ldots + x^2+x+1)$.

 

[Helly123]

Yes. Calculate the expressions:
$a_n = \frac{2}{3}a_{n-1}+\frac{1}{4}$ and $a_{n-1} = \frac{2}{3}a_{n-2}+\frac{1}{4}$ result in
$$a_n=\frac{2}{3}\left(\frac{2}{3}a_{n-2}+\frac{1}{4}\right)+\frac{1}{4}=\left(\frac{2}{3}\right)^2 a_{n-2}+\frac{1}{4}\left(\left(\frac{2}{3}\right)^0+ \left(\frac{2}{3}\right)^1 \right)$$
and in the next step with $a_{n-2}= \frac{2}{3}a_{n-3}+\frac{1}{4}$
$$a_n=\frac{2}{3}\left(\frac{2}{3}\left(\frac{2}{3}a_{n-3}+\frac{1}{4}\right)+\frac{1}{4}\right)+\frac{1}{4}=\left(\frac{2}{3}\right)^3 a_{n-3}+\frac{1}{4}\left(\left(\frac{2}{3}\right)^0+ \left(\frac{2}{3}\right)^1 + \left(\frac{2}{3}\right)^2 \right)$$
Now look at the pattern, go on until $a_n=\ldots \, \cdot a_1 + \ldots$, substitute $a_1=3$ and compute $\sum_{i=0}^n q^n$ with $q=\frac{2}{3}$. This sum can be calculated by $(x-1)^n=(x-1)\cdot (x^{n-1}+x^{n-2}+\ldots + x^2+x+1)$.

I will try to think on it

 

[Helly123]

You can't do this. For one thing, you can't have an index of $\infty$. You also can't have $\infty$ as part of an arithmetic expression like you have in $\infty - 1$.
fresh_42 and I are approaching this from two different directions: in post #23, he's working from $a_{n+1}$ down (i.e., starting with $a_{n+1}$ in terms of $a_n$, then with $a_{n+1}$ in terms of $a_{n-1}$, then with $a_{n+1}$ in terms of $a_{n-2}$, and so on. In post #5, I working my way from $a_2$ in terms of $a_1$, then to $a_3$ in terms of $a_1$, and then $a_4$ in terms of $a_1$. Do this as many times as necessary until you see a pattern.

Ok. I can't use $\infty$ on sequence formula because it just not appropriate mathematically?

 

[fresh_42]

Ok. I can't use $\infty$ on sequence formula because it just not appropriate mathematically?

Nobody uses this. It could only mean $a_\infty = \lim_{n \to \infty}a_n$ which is less ambiguous and more precise. There is no element $a_\infty$ in the sequence: each element of the sequence belongs to an index $n$. They move on to infinity, but each one of them is at the unique $n-$th place in the sequence. And if you will have your formula for $a_n$, then you cannot simply short cut the limit process by writing $\infty$, because there are no calculation rules for $\infty$. It might happen that it sometimes leads to the same result, but I wouldn't bet on it.

 

[Mark44]

Ok. I can't use $\infty$ on sequence formula because it just not appropriate mathematically?

Right. Writing $a_\infty$ might be forgiven as a sloppy way of writing $\lim_{n \to \infty}a_n$, but $a_{\infty - 1}$ has no meaning at all.

A misunderstanding you seem to have throughout this thread is what you're supposed to do, which is to find the limit of the sequence $\{a_n\}$ as n grows large. To do that, you need to get a formula for $a_n$ in terms of $a_1$ alone.
The problem states that $a_{n + 1} = \frac 2 3 a_n + \frac 1 4$ and that $a_1 = 3$.
One of your first steps was to rearrange the equation above, solving for $a_n$ in terms of $a_{n + 1}$. That provides you no help in finding the limit of the sequence. It also seems that you don't understand what a limit is or what is meant by the statement $\lim_{n \to \infty} a_n = L$. Being able to do this problem assumes that you have a good working knowledge of limits.

I've worked this problem, and found that the sequence has a limit somewhere between 0 and 1, but closer to 1. fresh_42 and I have given you some strong hints in post #5 (me) and post #23 (fresh_42). These are different approaches, but either one should lead to a solution.

 

[Dick]

To do that, you need to get a formula for $a_n$ in terms of $a_1$ alone.

That's not true. You can certainly find the limit of a sequence defined like this without finding a formula for $a_n$. Any reason why no one except me seems to be suggesting this way of doing it?

 

[Helly123]

Ok. How to do that?

 

[Mark44]

That's not true. You can certainly find the limit of a sequence defined like this without finding a formula for $a_n$. Any reason why no one except me seems to be suggesting this way of doing it?

I was able to find the limit of the sequence using the strategy I suggested, which is a variant of the technique that @fresh_42 suggested. You gave a hint of sorts in post #2. Can you elaborate a bit on what you said in that post?

 

[fresh_42]

I have a few doubts, that solving a differential equation is the appropriate way here.

 

[Dick]

Ok. How to do that?

Did you try following my suggestion in post 2? If $\lim_{n \to \infty}a_n = L$ then what is $\lim_{n \to \infty}$ of the equation $a_n = \frac{2}{3}a_{n-1}+\frac{1}{4}$??

 

[fresh_42]

Did you try following my suggestion in post 2? If $\lim_{n \to \infty}a_n = L$ then what is $\lim_{n \to \infty}$ of the equation $a_n = \frac{2}{3}a_{n-1}+\frac{1}{4}$??

Ah! Now I got it. I first thought you wanted to convert it into a differential equation and solve this.

Edit: But doesn't this leave us with the obligation to prove, that $L$ actually exists, since if we only get a statement: "If L then L=..."? And if we turn e.g. to Cauchy sequences to do so, we will arrive at the point, where the other methods started anyway.

 

[Helly123]

Did you try following my suggestion in post 2? If $\lim_{n \to \infty}a_n = L$ then what is $\lim_{n \to \infty}$ of the equation $a_n = \frac{2}{3}a_{n-1}+\frac{1}{4}$??

Ok.
What is it? Is there any clue for me?

 

[PeroK]

Ok.
What is it? Is there any clue for me?

To solve this problem you could a) work out what the limit must be and b) prove that rigorously. I think a) should be fairly straightforward. In general, for a recursive sequence of the form:

$a_{n+1} = f(a_n)$

You can, informally, say that if $a_n \rightarrow \ L$, then for "large enough" $n$ we have $a_n \approx L$. And, if $f$ is a continuous function we have:

$L \approx f(L)$

That seems like a general technique to get you started. And, you might like to think about how to prove this rigorously.

The other general technique is to look for monotone behaviour and boundedness. For example, you might be able to show something like:

$a_n > 0 \ \Rightarrow \ a_n > a_{n+1} > 0$

And, as any bounded monotone sequence has a limit, that would prove that a limit exists..

 

[Dick]

Ah! Now I got it. I first thought you wanted to convert it into a differential equation and solve this.

Edit: But doesn't this leave us with the obligation to prove, that $L$ actually exists, since if we only get a statement: "If L then L=..."? And if we turn e.g. to Cauchy sequences to do so, we will arrive at the point, where the other methods started anyway.

Sure, it does leave you with an obligation. I alluded to that in post 2. But it's straightforward to show that $a_n$ is decreasing and bounded below implying it does have a limit. I think that's much easier than actually finding an expression for $a_n$ and taking the limit.

 

[fresh_42]

But it's straightforward to show that anana_n is decreasing and bounded below implying it does have a limit.

Agreed. However, this uses a proposition from calculus I, whereas the computation can be done by school methods.

 

[Dick]

Agreed. However, this uses a proposition from calculus I, whereas the computation can be done be school methods.

Also agreed.

 

[Helly123]

Can i just get the solution please?
I have no idea what we discuss..

 

[fresh_42]

Can i just get the solution please?
I have no idea what we discuss..

Assume there is a limit $L := \lim_{n \to \infty}a_n$. Now what happens, it you take your definition $a_n=\frac{2}{3}a_{n-1}+\frac{1}{4}$ and go over to the limits on each side of the equation?

 

[Ray Vickson]

Can i just get the solution please?
I have no idea what we discuss..

NO: we are not allowed to give you the solution. That would go against the PF rules.

 

[Helly123]

I searched on internet.
Something says : if limit exist for sequences a_n. Then all the subsequences have the same limit as a_n

So.
Lim n->$\infty$ a_n = 2/3(Lim n->$\infty$ a_n ) + 1/4
So
1/3(Lim n->$\infty$ a_n ) = 1/4
Lim n->$\infty$ a_n = 3/4
Is it?

 

[PeroK]

I searched on internet.
Something says : if limit exist for sequences a_n. Then all the subsequences have the same limit as a_n

So.
Lim n->$\infty$ a_n = 2/3(Lim n->$\infty$ a_n ) + 1/4
So
1/3(Lim n->$\infty$ a_n ) = 1/4
Lim n->$\infty$ a_n = 3/4
Is it?

What else could it be?

 

[Helly123]

What else could it be?

I have no idea..
I just find it randomly.. and i don't even understand this is how it work actually

 

[fresh_42]

I searched on internet.
Something says : if limit exist for sequences a_n. Then all the subsequences have the same limit as a_n

So.
Lim n->$\infty$ a_n = 2/3(Lim n->$\infty$ a_n ) + 1/4
So
1/3(Lim n->$\infty$ a_n ) = 1/4
Lim n->$\infty$ a_n = 3/4
Is it?

Yes, although subsequence in this context isn't needed, simply the fact that $\lim_{n \to \infty}a_n = \lim_{n \to \infty}a_{n+k}$ for any finite $k$. I mean, we run to infinity, who cares where we start at? The problem with this solution is, that we still need an argument why $L$ exists at all. Not that we calculate with something, which doesn't exist. Such an argument can be given by the fact that $a_n < a_{n-1}$, and all are positive, so bounded from below by zero. Therefore a limit has to exist, which is a theorem from calculus: The sequence has to go somewhere and space is running out!

 

[Helly123]

Yes, although subsequence in this context isn't needed, simply the fact that $\lim_{n \to \infty}a_n = \lim_{n \to \infty}a_{n+k}$ for any finite $k$. I mean, we run to infinity, who cares where we start at? The problem with this solution is, that we still need an argument why $L$ exists at all. Not that we calculate with something, which doesn't exist. Such an argument can be given by the fact that $a_n < a_{n-1}$, and all are positive, so bounded from below by zero. Therefore a limit has to exist, which is a theorem from calculus: The sequence has to go somewhere and space is running out!

Hmm... i see... but. Its fun anyway. Thanks for all the disscussion before. :D