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I want to ask for two examples in the following cases:

1) Given a bounded sequence $\{a_n\}$, $$\lim_{n\to \infty}{(a_{n+1}-a_n)}=0$$ but $\{a_n\}$ diverges.

2) A function defined on real-line $f(x)$'s Taylor series converges at a point $x_0$ but does not equal to $f(x_0)$.

Thanks for your help.

Edit

in 2), I was thinking of the Taylor series of the function $f$ at the point $x_0$.

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    Is this homework? What have you tried?2011-03-27
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    @Qiaochu: not homework!2011-03-27
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    @Qiang Li: But how can a bounded sequence diverge?...2011-03-27
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    Question (2) could benefit from some rephrasing.2011-03-27
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    @Pacciu "Divergent" is the antonym of "convergent".2011-03-27
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    @Pacciu: other already gave examples! :)2011-03-27
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    Re the Edit of 2): The Taylor series of a function $f$ around a point $x_0$ does converge to $f(x_0)$ when evaluated at $x_0$. Always.2011-03-27
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    @Didier and Qiang: Oh, now I see... It's a matter of definitions. I say *$a_n$ diverges* iff $\lim a_n=\pm \infty$, while I use *$a_n$ oscillates* in case $\lim a_n$ does not exist. Thanks to both of you.2011-03-27
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    @Pacciu You should not. Honest.2011-03-27
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    @Didier: Why not? Your use of the verb *diverge* is typical of English mathematics; where I'm from this use is not common at all.2011-03-27
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    @Pacciu: "oscillates" is just one way in which a limit may fail to exist; and when the limit "equals $\infty$", the limit does not exist. Writing $\lim a_n=\infty$ does not mean the limit exists, it means the limit does **not** exist, and that the reason it does not exist is that the $a_n$ grow without bound. So it's wrong to say "$a_n$ oscillates" as a synonym of "$a_n$ does not converge" for the simple reason that it is false that the two are equivalent.2011-03-27
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    @Arturo: Mmmm... I'm still not convinced. The point is that your definition of "oscillating sequence" doesn' match mine. In fact, it seems to me that you call "oscillating" a *bounded sequence which does not converge*: e.g. $(-1)^n$ oscillate, but $(-2)^n$ don't. On the other hand, in my opinion $(-2)^n$ exhibits a clear oscillating feature, even if the amplitude of its oscillations increases without bound... But I think these are purely linguistic problems, not mathematical ones. P.S.: I never ever wrote $\lim a_n=\infty$ in my whole life; I think it is an ambiguous notation.2011-03-27
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    @Pacciu: Do you *have* a **formal** definition of "sequence oscillates"? Because the problem is that you are right now writing in English, and in English the terminology already exists and means something else entirely. You are free to call it whatever you want in your home, and to use standard nomenclature in your native language when speaking your native language, but if you try to export the nomenclature from one language to another you'll end up saying false things without meaning to. If I go around talking about "body extensions" (from French) I'm still speaking math nonsense in English.2011-03-27
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    @Arturo: In fact I'm not complaining, nor I'm trying to change the English usage of the verb *to diverge*; I was only trying to explain why I misinterpreted the text (answering to those *really nice people around* who do really like scoffing at newbies). So I don't see the point... BTW, *$a_n$ oscillates iff $\lim a_n$ does not exists*, while *$a_n$ diverges iff $\lim a_n=+\infty$ or $\lim a_n=-\infty$* (and in the first case, $a_n$ *diverges in the positive sense*, in the second it *diverges in the negative sense*) as I said before.2011-03-27
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    @Pacciu: And what does "the limit exists" mean? Usual definitions are such that if "$\lim a_n=+\infty$" holds, then the limit does *not* exist. E.g., standard definition is that the limit exists if and only there exists $L$ such that for all $\epsilon\gt 0$ there exists $N$ such that $n\geq N$ implies $|a_n-L|\lt\epsilon$. This means that even if you allow $L$ in the extended reals, $\lim a_n=+\infty$ implies the limit does not exist, so the conditions you give are not mutually exclusive...2011-03-27
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    ... And: *to diverge* comes from the Latin *devergere*, as you know, which (almost literally) means *to go in the opposite direction*. So, I cannot figure out why a *divergent* sequence can exhibit a feature like $(-1)^n$, which clearly doesn't go in any direction. XD2011-03-27
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    @Pacciu: And "passion" comes from the latin "to suffer", but when we use it when not speaking in Latin, it's not used to mean "suffering" (except in archaic phrases like "the passion of the Christ")2011-03-27
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    Oh, now I see the point... When you write $\lim a_n=+\infty$, this means the limit doesn't exist.... It is really amusing. So, tell me please: why do you write $\lim a_n$ in the LHside if the limit doesn't exist?2011-03-27
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    @Pacciu: As I said, this is specific notation that is explicitly introduced as shorthand for "the limit does not exist, and the reason it does not exist is..." Yes: it creates some confusion in students (but it is very useful shorthand in too many situations to abandon); I guess it's a question of choosing where and how you want to confuse people. (-:2011-03-27
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    @Arturo: In fact, this is what I'm saying... The usages may vary and sometimes misunderstandings may occur; it is natural. What it is unnatural here, among intelligent guys like the ones in this forum, is that someone can believe to be allowed to scoff at someone else only because of "linguistic" misunderstandings. For what is worth, I prefer to say that *the limit exists ad it is $\pm \infty$*, because this way I feel more comfortable when writing $\lim a_n$ in $\lim a_n=\pm \infty$. ;D2011-03-27
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    @Pacciu: I did not see "scoffing" (I hope I was not the one you perceived as scoffing at any rate). But under standard terminology, it would be incorrect in English to say what you prefer to say (see the definition I gave above for "limit exists"; it is simply the case that the limit does *not* exist in that case).2011-03-28
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    @Arturo: As I said, it's a linguistic issue, not a mathematical one. I do understand your point of view, and I will conform to your "rules" as long as I will post here. About the comments, never mind (the one I didn't like has been deleted); I'm sure the guys will understand and will try to be less sarcastic next times.2011-03-28

5 Answers 5

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Eric gave an example of 2). As regards 1), let $a_n=\cos(2\pi k/2^i)$ if $n$ is between $2^i$ and $2^{i+1}$ and $n=2^i+k$ with $0\le k\le 2^{i+1}-1$. Then $|a_{n+1}-a_n|\le2\pi/2^i\le4\pi/n$ hence $a_{n+1}-a_n\to0$ but the limit set of the sequence $(a_n)$ is $[-1,1]$.

Another example is $a_n=\cos(\log n)$. Then $|a_{n+1}-a_n|\le1/n$ and the limit set is $[-1,1]$.

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    I like this! Rotate around circle slower and slower, but not too slowly.2011-03-27
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    @Eric Yup. @Alon had the same idea.2011-03-27
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For 1): $$ 1,\frac12,0,\frac13,\frac23,1,\frac34,\frac24,\frac14,0,\frac15,\frac25,\frac35,\frac45,1,\frac56,\frac46,\frac36,\frac26,\frac16,0,\frac17,\dots $$ I leave it to you to find an explicit formula for $a_n$.

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    Very nice indeed!2011-03-27
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2) Define the function $f$ as follows: $$f(x)=e^{-1/x^2}\ \text{if } x>0$$ and $$f(x)=0\ \text{if } x\leq 0.$$ Now consider the Taylor series centered at zero. This provides an example of a non-analytic $C^\infty$ function. This taylor series converges everywhere, but is identically zero, and $f(x)$ is not identically zero.

Hope that helps,

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    @Eric: constant sequence of course converges!!2011-03-27
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    @Qiang: I realized before the comment! Fixed now :)2011-03-27
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    @Eric: now your sequence is not bounded. I also don't understand your second example. The Taylor series is identically zero, so the value of the Taylor series at $0$ agrees with the actual value, doesn't it?2011-03-27
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    @Eric: the sequence you gave is not bounded.2011-03-27
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    @Qiaochu @Qiang: My (1) is just not salvageable. I don't know what I was thinking... However (2) is correct (any point $x>0$ we have what was needed.)2011-03-27
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    @Erik: I was intended to require the Taylor series at the point $x_0$ too.2011-03-27
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    @Qiang: What do you mean? The Taylor series of $f$ centered at 0 converges at every real number. So choose any $x_0>0$, the series converges there but $f$ wont be zero. (And the Taylor series is identically the zero function if worked out)2011-03-27
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    To ask that the Taylor series of $f$ around $x$ does not converge to $f(x)$ when evaluated at $x$ could prove a daunting task, my friends...2011-03-27
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    @Eric: see my edit. I meant the Taylor series centered at $x_0$, in your language. :)2011-03-27
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    @Qiang: You can't of meant this? The Taylor series centered at $x_0$ always agrees with the function at the point $x_0$, by _definition_ of Taylor series. Didier hints at this in his comment.2011-03-27
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2) It is classical:

$$f(x)=\begin{cases} \exp (-\tfrac{1}{x}) &\text{, if } x>0 \\ 0 &\text{, if } x\leq 0 \end{cases}$$

and $x_0=0$.

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    the sequence is not bounded.2011-03-27
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    @cardinal: Baby Rudin. :)2011-03-27
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    @Qiang Li: I edited. I just misread the problem. ;D2011-03-27
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    @Qiang, you had the same comment, so I deleted mine.2011-03-27
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Hint for 1): Find a sequence $t_n$ which tends to infinity with $t_{n+1}-t_n \to 0$ as $n \to \infty$, and sample a periodic function such as $\sin(x)$ at the points $t_n$.