##\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.