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In Limit of $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$ 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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up vote 3 down vote accepted

(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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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-Alvarez Dec 28 '10 at 14:49

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 $1/2$.

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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
@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. – Arjang Dec 28 '10 at 13:06
@Arjang: but you need to settle on a definition of convergence before checking whether the sequence converges and computing its limit. – Mariano Suárez-Alvarez Dec 28 '10 at 13:59
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
@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-Alvarez Dec 28 '10 at 17:03

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.

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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. – Arjang Dec 28 '10 at 13:19
@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-Alvarez 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. – Arjang 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. – Arjang Dec 28 '10 at 14:31

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).

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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? – Arjang Dec 28 '10 at 13:22
@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-Alvarez 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-Alvarez 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? – Arjang 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. – Arjang Dec 28 '10 at 16:18

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