How can I find a starting point for problems with convergent subsequences?

[dmatador]

I am having a hard time finding a starting point for these problems. One is to find a sequence with subsequences that converge to 1, 2, and 3.

A similar problem (which would solve both problems) is to find a sequence that has subsequences that converge to every positive integer.

I am not so much looking for an outright answer, but how can I begin this?

 

[LCKurtz]

I am having a hard time finding a starting point for these problems. One is to find a sequence with subsequences that converge to 1, 2, and 3.

Think about doing something with every third term of your sequence. Then something else...

 

[Mark44]

Here's a sequence with subsequences that converge to 0 and 2.

{0, 3/2, 0, 7/4, 0, 15/8, 0, 31/16, ... }

Does that help you with your first problem?

 

[dmatador]

What is the explicit formula for that? I am getting too tired thinking about this for so long. I can see having two limits, but three?

 

[Dick]

I'll up the ante. Here's one with subsequences that converge to 1, 2 and 3. {1,2,3,0,0,0,...}.

 

[dmatador]

Do you not need an explicit formula? Does that even converge to anything but 0?

 

[Dick]

Do you not need an explicit formula? Does that even converge to anything but 0?

No, I'm wrong. I was thinking of series. Sorry. How about {1,2,3,1,2,3,1,2,3,...}.

 

[dmatador]

Sorry to nag, but you are saying that that is enough of an answer? What would the formula for that sequence be?

 

[Dick]

Sorry to nag, but you are saying that that is enough of an answer? What would the formula for that sequence be?

The sequence is pretty clear without an explicit formula isn't it? How about a_{3k}=1, a_{3k+1}=2, a_{3k+2}=3? You could probably also cook up something using (-1)^k or something, but what's the point? The sequence is {1,2,3} repeated. I think that expresses it more clearly than any formula.

 

[dmatador]

Yes, it is exceedingly obvious by looking at the first portion of the sequence written out, but I am just trying to be rigorous. Thank you.

 

[Dick]

Yes, it is exceedingly obvious by looking at the first portion of the sequence written out, but I am just trying to be rigorous. Thank you.

Have you thought about the second part? You can just use the same strategy of repeating stuff a lot. In a nonperiodic way. I wouldn't worry about the explicit formula. Just describe it.

 

[wisvuze]

for the first one, you can pick use a_n = 1 if n is a multiple of 10, a_n = 2 if n is even without being a multiple of 10, a_n = 3 if n is odd. 10x is unbounded, so you can find enough 1's to construct a subsequence of 1's and there are enough odds and evens to pick out enough 2's and 3's. I guess the general strategy is to pick "index patterns" for which starting a cycle through every natural number does not exhaust appearances of these patterns.

For the second one, you can use the same strategy as the first one

 

[dmatador]

Have you thought about the second part? You can just use the same strategy of repeating stuff a lot. In a nonperiodic way. I wouldn't worry about the explicit formula. Just describe it.

Would this work?

{.9, .99, .999, ... , 1.9, 1.99, 1.999, ... 2.9, 2.99, 2.999, ...}

 

[wisvuze]

Would this work?

{.9, .99, .999, ... , 1.9, 1.99, 1.999, ... 2.9, 2.99, 2.999, ...}

no that sequence isn't clear, are you saying it starts at .9 then you go infinity indices adding an 9 at the end everytime to get 1.. et c? how do you manage to make it to the 1.9 part? This could work i think , if you embed all of these terms together or distribute them somehow throughout the sequence so you can actually get "all of them" as you go to infinity, you just can't stack the sequences. You actually can't put them side by side either, because you'll have infinity many of the first .9, 1.9,2.9 kind .. et c

you can do that, or you can find some kind of identification between a positive integer and a big class of numbers, like if you thought about divisibilities ( or something )

 

[LCKurtz]

No, I'm wrong. I was thinking of series. Sorry. How about {1,2,3,1,2,3,1,2,3,...}.

Doesn't that qualify as working it for him??

 

[dmatador]

This sequence with subsequences converging to every natural number is definitely not as analogous as I thought it would be to the previous problem. I'll keep thinking about it, but if anyone has some helpful suggestions or hints then I would be glad to read them. Thank you for all the help thus far.

 

[wisvuze]

hint:
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10 11
1 2 3 4 5 6 7 8 9 10 11 12
... et c?

 

[dmatador]

I believe that I see what you are hinting at:

Is this legal, and if so, is it correct?

{1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, ... }

 

[wisvuze]

that's the same idea as what I hinted at, that one should work too; I just started with 10 numbers so
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2 , 3, 4 , 5 , 6 ,7 ,8 , 9, 10 , 11 ... } ..et c

do *you* think it's correct?

 

[dmatador]

Yes, although I still feel uncomfortable putting something down without knowing how to write a formula for it. I know that it isn't necessary, but it just seems a little more solid to me.

 

[Dick]

Doesn't that qualify as working it for him??

Yeah, sorry. But I thought Mark44's example was pretty close to showing it while being too complicated. I was more interested in what dmatador would do with the second part of the question after getting a simple example of the first.