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In Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$ Timothy Wagner gave a correct answer that was questioned for not having shown that the limit exists in the first place. My question is, are there any examples were a value of limit can be derived although the limit does not exist? Please note I am not questioning that limit must exist before it is exhibited but that, how a value for limit can be exhibited if the limit doesn't exist? Is that not a contradiction? On one hand we have a value for the limit and on the other hand the proof that it can't exist!

Please give an example were a more general form of convergence does not account for the calculated value. Otherwise it seems as if the value was hinting that the notion of convergence required adjustment and not the calculated value that required justification.

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jimjim
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4 Answers4

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Yes. Consider the sequence $a_{n+1} = 2a_n + 1$, where $a_0 = 0$. If this sequence converged to some value $L$, then $(a_{n+1})$ would converge to the same value, so we would get $L = 2L + 1$, which gives $L = - 1$. But the sequence actually diverges to infinity. I use this sort of example when I teach real analysis to drive home the point that we have to check for convergence before trying to compute the value of a limit.

This method also comes up in with infinite series. Consider the series $1 + -1 + 1 + -1 + \cdots$, which is not convergent. If it were convergent to a value $s$, then we would have $s = 1 + (-1 + 1 + -1 + \cdots)$ so $s = 1 - s$, which leads to $s = 1/2$. So we can compute a "value" even though the series diverges.

This fact is actually useful in practice when we study generalized notions of convergence. For the moment, define a notion of series convergence to be "nice" if, whenever a series $\sum a_n$ converges, we have $\sum a_n = a_0 + \sum a_{n+1}$ and $\sum -a_n = -\sum a_n$. Then if $1 + -1 + 1 + -1 + \cdots$ converges under any "nice" notion of convergence, its limit will be $1/2$.

Carl Mummert
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    The point here is that -1 is a repelling fixed point of the map z -> 2z+1. Considered as a Mobius transformation from the Riemann sphere to itself, there is a second fixed point at infinity which is attractive, which is why one sees this behavior when restricted to the real line; the sequence is trying to converge to its other fixed point, which doesn't exist in R. – Qiaochu Yuan Dec 28 '10 at 12:30
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    @Carl Mummert : The value of first example is justified using the Euler summability, so the fact that a values exists is not contradictory. The sequence might seem to diverge to infinity but that is due to the wrong notion of convergence being user, when the more generalized notion is used the value is well justifeid and very well might have been considered as an indicator that the notion of convergence needed to be upgraded. Again the second example is Cesàro summable and shows that a wrong notion of convergence is been refuted by the actual value of 1/2. – jimjim Dec 28 '10 at 13:06
  • @Carl Mummert:The part about generalized notion of convergence is interesting, is that the modern approach to summability? any book you would recommend?

    From the examples I conclude that in essence it is not the values that are not justified but the notion of convergence that was unjustifying them were not sufficient for the task at hand.

     – jimjim Dec 28 '10 at 13:09
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    @Arjang: but you need to settle on a definition of convergence before checking whether the sequence converges and computing its limit. – Mariano Suárez-Álvarez Dec 28 '10 at 13:59
  • @Mariano Suárez-Alvarez : But that is opposite of what I am trying to find out. There are plenty examples that show that something does not converge according to some definition and yet a value can be computed. But what they are really doing is that they are showing the computed value can not be accounted for a less than adequate definition of convergence for that example. A real example that a computed value is not the limit has to show that no matter how we define the convergence, the computed value will remain unjustified. I have edited the question to reflect this condition. – jimjim Dec 28 '10 at 14:11
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    Arjang: if you want to consider completely arbitrary modes of convergence, you could just define the sequence to converge to whatever value you want. So it is impossible to obtain an "example that a computed value is not the limit has to show that no matter how we define the convergence". – Carl Mummert Dec 28 '10 at 14:22
  • @Carl Mummert: If it is true that it is always possible to redefine the convergence to have every possible computed value as a limit, then why in cases that one computes a value, it is requested that they also show the limit exists? In case of examples you gave the computed limit was -1 or 1/2 , and there are convergence definitions that justified them, so what is the point by also requiring that sequence converges to those values? Those values not only show a limit exists but also provide the value. What is gained by requiring to show existence when the exact value of the limit is provided? – jimjim Dec 28 '10 at 15:55
  • Because we don't care whether the limit exists in some arbitrary mode of convergence: we care whether it exists in the particular mode of convergence we're studying. In real analysis, this means the usual $\epsilon/N$ definition of the limit. – Carl Mummert Dec 28 '10 at 16:14
  • @Carl Mummert : In that case, What is one to make of the computed values that have been obtained by using justifiable methods at every step? e.g. by the first example, one can justify the value being -1 , yet saying that the sequence diverges, so what is -1? is it not an acceptable limit value? if not then why not? Every step to obtain -1 was justified, then if it is the limit of the sequence why require existence? if it is not the limit of sequence, what is it? How could a value that is obtained by justifiable steps not be the answer to the limit? How to reconcile "divergent" and "-1"? – jimjim Dec 28 '10 at 16:33
  • @Arjang: in that example, one is not justified in talking about the limit $L$ of sequence unless one assumes it exists. If you do not know that something you intend to denote $L$ exists and is a real number, you simply cannot consider the number $L+1$: it simply does not make sense! – Mariano Suárez-Álvarez Dec 28 '10 at 16:37
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    @Mariano: Dear Mariano and Carl, Arjang is not (as far as I can tell) suggesting that one can define the limit to be "whatever value you want", but is pointing out that there are various natural extensions of the notion of convergence, which satisfy the usual rules for computing limits, and which apply to the examples in Carl's answer. At least in the most recent version of the question, what Arjang is asking for is example of expressions for which one can compute an apparent value of the limit, but which don't actually have a limit with respect to any reasonable generalization of the ... – Matt E Dec 28 '10 at 16:53
  • @Mariano Suárez-Alvarez : I used to think so, but now I have come to question it. It used to seem natural, but I can't see what is wrong with not doing so. If I can see that it causes contradiction then I go back to the old belief gladly. But everything that I used to think is the example of why it is the case has turned out to be just deficiency of how the convergence was defined. That is why I asked the main question. I know it seems crazy, but without seeing the contradiction it causes I don't know how to dismiss it. Every example where existence was assumed has ended up justified. – jimjim Dec 28 '10 at 16:54
  • ... notion of convergence. (At least, this was my reading of the question.) Regards, – Matt E Dec 28 '10 at 16:54
  • @Matt E, thank you, you said it better than I could. – jimjim Dec 28 '10 at 16:56
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    @Matt: the existence of sequences to which different reasonable notions of convergence attach different limits follows from the result of Lorentz I mentioned in my answer (the description of the domain of coincidence of all Banach limits is sufficiently simple that I am pretty sure one can explicitly construct a sequence which does not belong to it) – Mariano Suárez-Álvarez Dec 28 '10 at 17:03
  • By the way, the series $1-1+1-1\dots$ is in the domain of coincidence of all Banach limits, as all periodic sequences. – Mariano Suárez-Álvarez Dec 28 '10 at 17:09
  • @Mariano: Dear Mariano, Thanks for your reply. (I hadn't read your answer when I wrote my comment, and then, after I read it, I realized that it anticipated and answered my comment. Sorry about that!) Best wishes, – Matt E Dec 28 '10 at 19:47
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(This should be a comment under Carl's answer.)

@Arjan: One very general way in which to formalize your idea of "possible definitions of convergence" is that of Banach limits. Indeed, a Banach limit is essentially a convergence notion for bounded sequences satisfying a minimal set of reasonable conditions.

Banach limits do exist---this was proved by Banach himself as a way to show off his Hahn-Banach theorem---and all Banach limits agree with the usual limit on sequences that converge. More interestingly, the set of bounded sequences on which all Banach limits agree is strictly larger than the set of convergence sequences, and it was determined explicitly by [Lorentz, G. G. A contribution to the theory of divergent sequences. Acta Math. 80, (1948). 167--190. MR0027868]

Now, the result of Lorentz implies that not all Banach limits agree on all sequences (in other words, that there is more than one Banach limit) It follows from this that there are divergent sequences which can be assigned at least two different limits under different extensions of the notion of convergence.

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Mariano Suárez-Álvarez
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  • Lorentz's paper is very beautiful and readable (and the typography is very nice!). The result I quote is just his Theorem 1, but the paper goes on with nice statements... – Mariano Suárez-Álvarez Dec 28 '10 at 14:49
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What he showed was that if a limit $L$ exists, then it has to satisfy the equation $L = \sqrt{7+L}$. This is typical of exercises early in a real analysis class where the only convergence theorem you have available is that bounded monotone sequences converge (this is easy to show given that the reals have the least upper bound property).

So if we have a sequence $(a_n)$ that is defined recursively by $a_1 = x$, $a_{n+1} = f(a_n)$ and is monotone and unbounded, then any fixed point of $f$ is going to be a candidate for a limit, but $(a_n)$ doesn't converge.

kahen
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  • There is a hidden assumption, that if there is a value that satisfies the equation then the notion of convergence has already accounted for it. A classic example often seen is Grandi's series that by the primitive notion of convergence is suppose to not converge, making the vale 1/2 to seem unjustified, where as once the notion of convergence is upgraded the value is justified and it becomes obvious initially the wrong notion of convergence was being used and indeed the value of 1/2 is more meaningful than what was considered not to be convergent. – jimjim Dec 28 '10 at 13:19
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    @Arjang: but there are other ways of "upgrading" the notion of convergence that make the series converge to other numbers... So it is difficult to see how significant is your last sentence. – Mariano Suárez-Álvarez Dec 28 '10 at 14:01
  • @Mariano Suárez-Alvarez : That is the point, then one can very well argue that it was not the computed value that was wrong but the notion of convergence that was used. In that case what is really being justified? That the computed value can not be accounted for by the primitive notion of convergence or, that the value is not justified at all? If the latter then every time a value is computed one must show that the limit did exist, otherwise the limit does exist and is given by the value but the notion of convergence was not adequate, so no need to show that limit existed, value is the limit. – jimjim Dec 28 '10 at 14:16
  • What I am after is an example where there are no known generalized convergence definitions that can account for a computed value. The requirement of "show a limit exists before saying this is the value of the limit" instead of "here is a value hence the limit exists but what is the convergence definition that would make it valid? Were no known generalized convergence definition will justify the value" is what I am after. Any guidance on how to clear up, rephrase the question are welcome. – jimjim Dec 28 '10 at 14:31
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The point is often when we try to derive the limit, we invoke algebraic manipulations which would only work if it has already been established that a limit exists (say, by showing the sequence is cauchy).

Dactyl
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  • The question was why is that so? That is why when a value is been exhibited, then is it not the proof that limit exists? and is given by the value? – jimjim Dec 28 '10 at 13:22
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    @Arjang: by the very simple reason that you just cannot exhibit something which does not exist. As others have told you already, what you call "exhibiting the value $\alpha$ of the limit" is more exactly "proving the (conditional) statement «if the limit exists, then its value is $\alpha$»". – Mariano Suárez-Álvarez Dec 28 '10 at 14:37
  • In particular, in all cases where you "exhibit the value of the limit" you do so by using the fact that, being the limit, it is related in certain specific ways to the sequence---this is what justifies the manipulations you do. But that fact depends on its being the limit of the sequence in a specific sense, whatever it may be! – Mariano Suárez-Álvarez Dec 28 '10 at 14:52
  • @Mariano Suárez-Alvarez : But it seems the justification of manipulation and the existence of value are separate. if a value $\alpha$ can be explicitly given, then isn't that proof that it exits? What is more powerful method of showing existence of something: showing the thing itself or showing the thing has to exist? – jimjim Dec 28 '10 at 16:17
  • @Mariano Suárez-Alvarez : Let's simplify the question, What is wrong in the logic of :"This is how one can calculate $\alpha$, there is no need to show that $\alpha$ exists to begin with" (because if $\alpha$ didn't exist then what is computed and exhibited? assuming the computation of $\alpha$ is according to rules. – jimjim Dec 28 '10 at 16:18
  • What's wrong is that that does not make sense. You can only "compute a value" if that value is defined in someway. I don't think I can help you clarify this: IMO this is best done actually talking with someone. – Mariano Suárez-Álvarez Dec 28 '10 at 16:24
  • @Mariano Suárez-Alvarez : Thank you, I have consulted my analyst (it is good to have friends in math dept ). But a computed value that is found by using consistent methods and does not cause a contradiction must be accepted as the value of the limit. The fact that the a specific definition of the limit does not account for it is an indication of the definition not being general enough. Showing that a limit exist and then find it, is actually showing that limit exists according to some definition and not that it exists independently of the definition being used that it can exist. – jimjim Dec 30 '10 at 23:35