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Is this a valid sequence: $$\left\{\frac1{(n-3)}\right\}?$$ That is, can a sequence have individual terms that are undefined?

If so, does this mean that the above sequence is unbounded (since the third term is not smaller than any real number)?

[What I do know is that if the above sequence were valid, it would be convergent (by definition), which would in turn mean that it is bounded (by the Boundedness Theorem).]

Subsequent to the main discussion here, I have just come across an old exam question that asks us to show that $\langle\tan(\frac{\sqrt n\pi}4):n∈N\rangle$ is divergent. But this sequence is undefined at say, $n=4$, so like the one above, it isn't even a valid sequence, is it? (I've used a different notation for sequences on Brian M Scott's advice.)

Xander Henderson
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ryang
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    Actually, one way to define a sequence is as a function from $\mathbb N$ to the set in which the sequenc terms live, e.g. $\mathbb R$ in your case. Therefore, every term of the sequence must be defined. – Johannes Kloos May 11 '12 at 06:42
  • Why would you want such a thing?
    (I mean undefined elements, not helpfulness, of course.)     – copper.hat May 11 '12 at 06:57
  • So, the boundedness/unboundedness of a sequence is relevant only when n tends to infinity, right? Also, am I correct to understand that for sequences, boundedness<=>existence of limit ? (This last question is actually the motivation for my main question above. :) – ryang May 11 '12 at 07:02
  • $0,1,0,1,0,1,0,1,\dots$ is bounded but does not converge. It has two limit points but no limit. – Gerry Myerson May 11 '12 at 07:08
  • @Gerry Oh, my understanding was that sequences like sin x and the one you gave have multiple limits. Ok, then am I correct to understand that boundedness<=>existence of limit points ?? p.s. I'm currently doing a first course in Analysis. – ryang May 11 '12 at 07:11
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    @Ryan: Boundedness implies the existence of limit points (this is the Bolzano-Weierstrass theorem), but not conversely: 0,1,0,2,0,3,... is not bounded, but has 0 as a limit point. – Johannes Kloos May 11 '12 at 07:17
  • @Johannes Oh. So having a unique limit point doesn't even necessarily imply the existence of a limit! Wow, great example. Thanks! – ryang May 11 '12 at 08:50
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    @Ryan: $+\infty$ is also a limit point of Johannes Kloos's sequence. Situations like this are one of the many reasons why extending the real numbers to include one or more points 'at infinity' is a very good idea. –  May 12 '12 at 02:24
  • @Hurkyl Yes, I already saw that. Basically it's to do with how we have defined "limit". I used to think that "$\lim_{n \to c} f(n)=\infty$" is perfectly uncontroversial, but since this discussion I have decided to try to remember to write "$f(n)\to \infty$ as $n\to c$" instead when referring to an infinite limit point. Same for multiple limit points: I also thought that "$\lim_{n \to c} f(n) = 1,5$" was perfectly meaningful, but I have also since learnt better! – ryang May 12 '12 at 09:26

3 Answers3

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No one has yet actually answered your first question. The real problem with $\left\{\frac1{n-3}\right\}$ is simply that it's very sloppy notation. However, it will normally be understood to mean $$\left\langle\frac1{n-3}:n\in\Bbb N\text{ and }n>3\right\rangle\;,$$ which is a perfectly good sequence with domain $\{4,5,6,\dots\}$ instead of $\Bbb N=\{0,1,2,\dots\}$. As such it is bounded and converges to $0$.

If one attempts instead to interpret it as $$\left\langle\frac1{n-3}:n\in\Bbb N\right\rangle\;,$$ a sequence with domain $\Bbb N$, it is of course nonsense.

Any ordered set that 'looks like' $\Bbb N$ can reasonably be used as the domain of a sequence, but in practice you'll find that the domain is almost always of the form $\{n\in\Bbb Z:n\ge n_0\}$ for some integer $n_0$. Note that $n_0$ isn't always non-negative: you may find sequences defined on domains like $\{-1,0,1,2,\dots\}$.

By the way, if you must abuse the curly braces by using them for sequences, and unfortunately it's a very common practice, you should probably indicate the limits of the domain, e.g., $$\left\{\frac1{n-3}\right\}_{n=4}^{\infty}\qquad\text{or}\qquad\left\{\frac1{n-3}\right\}_{n\ge4}\;.$$

The matter of limit points of sequences is complicated by inconsistent terminology. It makes sense to talk about the limit of a sequence only when the sequence converges to some number $L$, in which case all of its subsequences converge to $L$ as well. If the sequence does not converge, I prefer to talk about its cluster points, to distinguish them from the unique limit of a convergent sequence; these are the points to which some subsequence of the given sequence converges. A bounded sequence of real numbers always has at least one cluster point, and examples like $\langle (-1)^n:n\in\Bbb N\rangle$ show that it may have more than one. It's not hard to show that if a bounded sequence has only one cluster point, the sequence actually converges to that cluster point, which is therefore the limit of the sequence. The example $$x_n=\begin{cases}0,&\text{if }n\text{ is even}\\n,&\text{if }n\text{ is odd}\end{cases}$$ given by Johannes Kloos, however, shows that this does not necessarily hold for unbounded sequences: this has the unique cluster point $0$ but obviously does not converge to $0$.

Brian M. Scott
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  • Thanks for your detailed answer. Can you help me check if this understanding is correct: [bounded + unique limit point] <=> [limit exists] ? – ryang May 11 '12 at 08:35
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    @Ryan: Yes. For a bounded sequence, the largest limit point is the limes superior and the smallest limit point the limes inferior. If they coincide, they are the limit of the sequence. – Michael Greinecker May 11 '12 at 08:39
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    @Brian I just came across an old exam question that asks us to show that $\left\langle tan(\frac{\sqrt n\pi}4):n\in\Bbb N\right\rangle;$ is divergent. But this sequence is undefined at say, $n=4$, so it isn't even valid, is it?? Do you think this sequence is as nonsensical as $\left\langle\frac1{n-3}:n\in\Bbb N\right\rangle;$ ? – ryang May 12 '12 at 01:23
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    @Ryan: It's worse: the $1/(n-3)$ sequence requires only a minor adjustment in the domain, to $\Bbb N\setminus{0,1,2,3}$, while the tangent sequence requires deleting from $\Bbb N$ the even squares. – Brian M. Scott May 12 '12 at 04:26
  • @Brian Gracias! I really appreciate and marvel at this community/website. It's so immensely helpful and I have honestly learnt more here than from my instructors. – ryang May 12 '12 at 09:34
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A sequence is, as Johannes Kloos pointed out, usually defined to be a function that has the natural numbers as their domain. There are cases where one has to rigidly follow this definition. When working with sequence spaces, on might have to evaluate things like $\sum_{n=1}^\infty |x_n|$ which only make sense if the sequence is defined for every coordinate.

What is however the case, is that for many problems, you can ignore many terms of the sequence. If you take some well defined sequence of real numbers and change finitely many terms of the sequence, you will get a sequence that shares some of the properties of the old sequence but, obviously, not all of them. "Asymptotic" properties do not change: The sequence will have the same limit points, the lim inf and the lim sup will be the same, it will be bounded just in case the old sequence was, if it converges, it will converge to the same point...

Similarly, some properties change and some properties remain the same when "shifting" coordinates. That is, if your original sequence is $(x_n)_{n=1}^\infty$ you can create a new sequence $(x_y)_{n=1}^\infty$ by picking some natural number $m$ and letting $y_n=x_{n+m}$ for all $n$. This sequence will again share these asymptotic properties above.

Now if you have a sequence that is not defined for finitely many coordinates, you can use these two methods described above to create another sequence and study the asymptotic properties of the latter. Since it doesn't matter how exactly you modified the original sequence, you can say, without running into any problems, that your old sequence had already these asymptotic properties. So a non-well-defined "sequence" can meaningfully be said to converge. If the "sequence" starts with a negative number, shifting can transform it into an ordinary sequence.

Note: Experienced mathematicians will usually do these things automatically without worry. But your teacher in elementary courses might well demand that your sequences are well-defined functions with domain $\mathbb{N}$, so you will have to transform your improper sequence into a proper one yourself first.

Michael Greinecker
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The definition of "sequence" should not be too stiff. A good compromise is the following.

Definition. Let $X$ be a set. A sequence in $X$ is any function defined on a subset of $\mathbb{N}$ that contains all but finitely many elements of $\mathbb{N}$, and taking values in $X$.

This definition is very good for the theory of convergence, and covers all the elementary situations. The assumption that the domain of the sequence must be $\mathbb{N}$ is clearly a useless obstruction.

Siminore
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