##\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.
  • #36
Helly123 said:
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}##??
 
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  • #37
Dick said:
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.
 
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  • #38
Dick said:
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?
 
  • #39
Helly123 said:
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..
 
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  • #40
fresh_42 said:
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.
 
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  • #41
Dick said:
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.
 
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  • #42
fresh_42 said:
Agreed. However, this uses a proposition from calculus I, whereas the computation can be done be school methods.

Also agreed.
 
  • #43
Can i just get the solution please?
I have no idea what we discuss..
 
  • #44
Helly123 said:
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?
 
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  • #45
Helly123 said:
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.
 
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  • #46
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?
 
  • #47
Helly123 said:
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?
 
  • #48
PeroK said:
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
 
  • #49
Helly123 said:
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!
 
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  • #50
fresh_42 said:
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
 
  • #51
fresh_42 said:
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!
yes i agree
 
  • #52
fresh_42 said:
Y 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!
It's even more useful to consider what happens if an = L+x. That will both give you a proof of the existence of the limit and an explicit formula for an as well.
 
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