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

In summary, the problem involves finding the formula for the sequence ##a_n## given ##a_1 = 3## and ##a_{n+1} = \frac{2}{3}a_n + \frac{1}{4}##. The sequence was found to be neither arithmetic nor geometric. The suggested approach was to expand the terms using function notation and continue until a pattern for ##a_n## can be identified. The final step would be to take the limit as ##n## approaches infinity to determine the limit of the sequence.
  • #1
Helly123
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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?
 
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  • #2
Helly123 said:

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?
 
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  • #3
Helly123 said:

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 ?##
 
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  • #4
Helly123 said:
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##.

fresh_42 said:
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.
 
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  • #5
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##.
 
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  • #6
Mark44 said:
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.
 
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  • #7
fresh_42 said:
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}##.
 
  • #8
Mark44 said:
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.
 
  • #9
Dick said:
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## ?
 
  • #10
Helly123 said:
##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##.
Helly123 said:
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.
Helly123 said:
And take the limit as x approaches ##\infty## ?
No, as n approaches ##\infty##.
 
  • #11
fresh_42 said:
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 said:
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##
 
  • #12
Helly123 said:
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.
 
  • #13
Sorry. IT HAS TO BE ##\lim_{n \to \infty} ## ALL THIS TIME. NOT ##\lim_{x \to \infty} ## . i made mistake in the question
 
  • #14
Helly123 said:
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.
 
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  • #15
Mark44 said:
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##
 
  • #16
Helly123 said:
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.
Helly123 said:
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.
Helly123 said:
Usually taking the limit as x approaches ##\infty## , i have to substitute the x variable into ##\infty##
There is no x in this problem.
 
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  • #17
Helly123 said:
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.
 
  • #18
Helly123 said:
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##.
 
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  • #19
fresh_42 said:
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}##

fresh_42 said:
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
 
  • #20
Helly123 said:
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)##
 
  • #21
Mark44 said:
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?
 
  • #22
fresh_42 said:
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
 
  • #23
Helly123 said:
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.
 
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  • #24
Helly123 said:
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.
 
  • #25
fresh_42 said:
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?

$$
a_\infty = \frac{2}{3}a_{(\infty-1)}+\frac{1}{4} $$
 
  • #26
Helly123 said:
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.
 
  • #27
Helly123 said:
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)##.
 
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  • #28
fresh_42 said:
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
 
  • #29
Mark44 said:
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?
 
  • #30
Helly123 said:
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.
 
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  • #31
Helly123 said:
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.
 
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  • #32
Mark44 said:
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?
 
  • #33
Ok. How to do that?
 
  • #34
Dick said:
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?
 
  • #35
I have a few doubts, that solving a differential equation is the appropriate way here.
 
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