What does the limit imply here?

  • Thread starter Thread starter hassman
  • Start date Start date
  • Tags Tags
    Limit
AI Thread Summary
The discussion centers on the implications of a limit statement in relation to sequences of natural numbers. It clarifies that while a sequence can be unbounded, indicated by the limit approaching infinity, this does not necessitate a strict order or uniformity in the sequence's values. Specifically, the limit does not imply that each term must be greater than the previous one for all n, but rather that there exists a point beyond which the sequence consistently increases. Additionally, the magnitude of the jumps between terms can vary significantly. Overall, the limit statement allows for flexibility in the sequence's structure while ensuring it ultimately grows without bound.
hassman
Messages
36
Reaction score
0
S'ppose this statement:

p_1, p_2, p_3,... \in \mathbb {N}

I do understand that p-series is infinite (from the dots) and that every p from the series is a natural number.

However, does the statement also imply that there is no particular order in the series? I.e. is it possible that

p_1 = 3,<br /> p_2 = 66,<br /> p_3 = 1

Does the above statement imply that there is no restriction that some p or even all of them are equal? I.e.

p_1 = 3, p_2 = 3, p_3 = 12


If all of the above is true, then what does this mean:


p_1, p_2, p_3,... \in \mathbb {N}

<br /> \lim _{n \to \infty} p_n = \infty

Does the addition of limit statement imply some sort of order in the series?

Oh, and how do I make new line in latex? \\ and \newline don't seem to work.
 
Mathematics news on Phys.org
Yes, you understood it correctly. Such a list of numbers p is what we usually call a sequence, and indeed its elements can be anything.

In special cases, sequences may have a limit. For example, if
\lim_{n \to \infty} p_n = L
then we mean that if we make n larger and larger, p(n) will get closer and closer to L. Of course, if p(n) can only take integer values, this means that all the p(n) are equal to L for n sufficiently large. If you substitute L for infinity, we mean that the sequence is unbounded. You should really see this as a convention, although it looks a bit like a limit: we can get the values of the sequence "closer and closer to infinity" as n gets bigger and bigger. More correctly: if we make n bigger and bigger then p(n) will get bigger and bigger -- conversely: we can get p(n) bigger than any number we want by choosing n sufficiently large.

Note that
(1): \lim_{n \to \infty} p_n = \infty
and
(2): "the limit does not exist"
are two very different statements. For example, the sequence
p_n = n
satisfies (1), while
p_n = (-1)^n
satisfies (2).
 
Ok, so the limit statement does not imply that p(1)=1, p(2)=2, etc. In other words it does not imply a particular sequence, just a sequence where p(n) gets larger as n gets larger.

I just want to know what the limit statement in combination with the definition of sequence eliminates. Does it eliminate that for some n p(n)>p(n+1)? Does it eliminate that for some n p(n) = p(n+1)?

If I understood correctly, the limit statement implies that p(n+1) > p(n) for every n, right? however it does not imply what the jump is from p(n) to p(n+1). it could be that p(1) = 1, p(2) = 13, p(3) = 1000, right? That the sequence is indeed increasing but that the magnitude of each jump is not fixed. Is that right?
 
hassman said:
Ok, so the limit statement does not imply that p(1)=1, p(2)=2, etc. In other words it does not imply a particular sequence, just a sequence where p(n) gets larger as n gets larger.
Indeed.

hassman said:
I just want to know what the limit statement in combination with the definition of sequence eliminates. Does it eliminate that for some n p(n)>p(n+1)? Does it eliminate that for some n p(n) = p(n+1)?
Neither. If you give me some integer n, I can always construct a sequence going to infinity and having p(n) > p(n + 1) for that specific n (just define p(k) = k for all k not equal to n, and p(n) = n + 2).

hassman said:
If I understood correctly, the limit statement implies that p(n+1) > p(n) for every n, right?
Not for every n. It does imply that there is some (possibly extremely large) number N, such that p(n + 1) > p(n) whenever n > N (i.e. from a certain point the sequence must be increasing). In fact, that's very nearly the definition:
\lim_{n \to \infty} p(n) = \infty
means that for any L there exists N, such that
whenever n > N, p(n) > L.
 
hassman said:
Ok, so the limit statement does not imply that p(1)=1, p(2)=2, etc. In other words it does not imply a particular sequence, just a sequence where p(n) gets larger as n gets larger.
Be careful here. If \lim_{m\rightarrow \infty} p_n= \infty that means p_n eventually becomes larger than any given real number. It does NOT mean that it does that in any simple way! For example, p_n= n for n even, p_n= n-2 for n odd gives the sequence -1 2, 1, 4, 3, 6, 5, 8, 7, etc. That "goes to infinity" but I don't think we would say "p_n gets larger as n gets larger" since for every even n, the next number is smaller.

I just want to know what the limit statement in combination with the definition of sequence eliminates. Does it eliminate that for some n p(n)>p(n+1)? Does it eliminate that for some n p(n) = p(n+1)?
No, neither of those things.

If I understood correctly, the limit statement implies that p(n+1) > p(n) for every n, right?
No, it does NOT! Saying "p_{n+1}&gt; p_n" for all n is simply saying that \{p_n\} is an "increasing" sequence which may go to infinity or converge to a finite number.

however it does not imply what the jump is from p(n) to p(n+1). it could be that p(1) = 1, p(2) = 13, p(3) = 1000, right? That the sequence is indeed increasing but that the magnitude of each jump is not fixed. Is that right?
 
HallsofIvy said:
No, it does NOT! Saying "p_{n+1}&gt; p_n" for all n is simply saying that \{p_n\} is an "increasing" sequence which may go to infinity or converge to a finite number.

As an example of the latter, consider the following sequence of rational number (I don't think an example exists for natural numbers);
p_n = 1 - 1/n
The first terms are then 0, 1/2, 3/4, 7/8, ...
The sequence always increases, but the limit is 1, not infinity.
Also note that the limit is a "real" limit in the sense that there does not exist an n such that pn = 1. However, by definition of the limit, you can get arbitrarily close: if you tell me how close you want to get to 1 (for example: within 0,001) I can give you an n which realizes that, i.e. pn will be and remain that close to the limit 1 (for 0,001, any n bigger than 1000 will do). In case the "limit" is infinity you need to replace "close to the limit" by something more sensible that expresses that we mean: the sequence is unbounded. In fact, the exact formulation is then: if you tell me how large you want the sequence to get (for example, bigger than 1000000) I can give you an n such that pn is bigger than and remains bigger than 1000000.
 
Thanks a lot guys, very clear explanations.
 
Back
Top