# What does the limit imply here?

1. Jun 6, 2009

### hassman

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

$$p_1 = 3, p_2 = 66, p_3 = 1$$

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:

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

$$\lim _{n \to \infty} p_n = \infty$$

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.

2. Jun 6, 2009

### CompuChip

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).

3. Jun 6, 2009

### hassman

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?

4. Jun 6, 2009

### CompuChip

Indeed.

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).

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.

5. Jun 6, 2009

### HallsofIvy

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.

No, neither of those things.

No, it does NOT! Saying "$p_{n+1}> p_n$" for all n is simply saying that $\{p_n\}$ is an "increasing" sequence which may go to infinity or converge to a finite number.

6. Jun 6, 2009

### CompuChip

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.

7. Jun 6, 2009

### hassman

Thanks a lot guys, very clear explanations.