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[update 3]: This question was stated from a wrong premise ("...fails obviously...") which I became aware of after comments and answers, and thus I tend to retract it. But there are those constructive answers, sheding light on this, so I think it is better to keep the question together with the answers alive

This is merely an accidental question, to improve my understanding of the concept of divergent summation.

I'm nearly completely used to the assumtion, that $ \sum_{k=0}^\infty x^k = {1 \over 1-x}$ by analytic continuation can be inserted in any formula (except of course for $x \ne 1$) - maybe I'm so-to-say over-experienced by sheer practice. On the other hand I think to remember to have read in Konrad Knopp's book, that divergent summation in the case of geometric series can be inserted in every analytical expression (I'll to check this possibly false memory when I've Knopp's book available again).

But here is an example, where the identity fails obviously:
$$ e^{-1-2-4-8-16- \cdots }={1 \over e^1}{1 \over e^2}{1 \over e^4} \cdots \ne e^1 $$

How can I characterize the range of algebraic operations, where such (even much standard) analytic continuation is applicable and where not? (Other examples might be the insertion of $\zeta$-values at negative arguments in place of their sum/product-representations in algebraic formulae)

[remark in the update 3]: analytic continuation needs some variable parameter with a possible value for which the expression is true/convergent. It proceeds in that that parameter gets changed as far the expression is analytic and convergent - and then analytic continuation is tried by further operations, coordinate change and shift of the range for the parameter. In the above formula such a variable parameter should be included, say the base parameter for the geometric series should be kept variable and for this the analytic continuation should then be attempted. This is kindly reflected in R.Israels answer

[update 2]: I'll add some context for this question from my comment to R. Israels's answer. It should shed much mor light on the intention of my question:
My question arose today when I re-read an older discussion of mine in the tetration-forum, where I didn't find an answer nor even a suitable direction for an answer (for my level of understanding at that time) Here is the link to the discussion where I had posed this in context with iteration-series and had already landed at that example of my current question: .

I've done a quick view into the chapter "divergent series" in Knopp's monography (in german language). I see at least one formulation which I might have overgeneralized and not taken precisely enough. I'll paraphrase it here to show the root of my concern:

(chap XIII, par. 261.) (...) "in a reasonable way" - this could also be interpreted, that we assign the sequence (s_n) in such a way a value s, that, wherever this sequence appears in a formula as a result of a computation, we should assign that value s always or at least generally to that result (...) (par 262.) (...) Whether now, whereever this series $\sum (-1)^n $ occurs as a result of a computation, we should assign it the value $\frac 12$ - this cannot be decided without further consideration. With the representation $ {1 \over 1-x} = \sum x^n $ for $x=-1$ this however is surely the case. (...)

It seems, I took that remarks too wide when I studied this chapter, and got too much unsensitive against the geometric series $\sum 2^k$ and its relatives... possibly I should be more critical today even to Knopp's formulation, which seems a bit too vague in the light of my concern today.

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What do you mean when you say "by analytic continuation can be inserted to any formula?" You cannot analytically continue the sum in question beyond the unit disk. Afterall, an analytic function needs to be finite everywhere and it's clearly not at $x=1$. – Alex R. Dec 10 '12 at 22:13
You should not "insert in any formula" unless you know a LOT more about analytic continuation. Why do you say your example "obviously fails" ... you did not "apply analytic continuation" to it. – GEdgar Dec 10 '12 at 22:18
@Alex: hmm, I cannot well answer your comment at the moment and thus improve my question at the moment. There's somehow knotted spaghetti in my thoughts - I think it's better I'll look at Knopp's book first to get thoughts straight... – Gottfried Helms Dec 10 '12 at 22:20
@GEdgar: well, I misread your comment at the first time, but after R. Israel's answer I think I got now clear, what you had in mind. Thanks for the reminder to be more careful when getting triggered to conjectures by "obvious" symbolic notations when deep complicated things like analytic continuation are involved... – Gottfried Helms Dec 11 '12 at 21:42
up vote 5 down vote accepted

I don't know what is your criterion for validity of the insertion. Let's say $f$ is an analytic function on a domain $D$, and the series $\sum_j g_j$ converges for $z$ in a domain $W$ to a function $g$ that has an analytic continuation to a larger domain $U$, with $g(U) \subseteq D$. Then $f(\sum_j g_j(z))$ for $z \in W$ has an analytic continuation to $f(g(z))$ on $U$. Is that what you're thinking of? In your example, with $f(z) = \exp(-z)$ and the series $\sum_{j=0}^\infty z^j$, $g(z) = 1/(1-z)$, it is indeed true that $$\exp\left(-\sum_{j=0}^\infty z^j\right) = \prod_{j=0}^\infty e^{-z^j}$$ has an analytic continuation to $\exp(-1/(1-z))$ on ${\mathbb C} \backslash \{1\}$, with value $e$ at $z = 2$. However, I would avoid writing this as $$ \prod_{j=0}^\infty e^{-2^j} = e$$

EDIT: Note also that it is possible to have $f(\sum_{j=1}^N c_j z^j)$ converge uniformly on compact subsets of domain $A$ to an analytic function $g(z)$ and uniformly on compact subsets of domain $B$ to a different analytic function $h(z)$ where $g(z)$ and $h(z)$ are not analytic continuations of each other. This will happen if $\sum_j c_j z^j$ has a finite nonzero radius of convergence and $f$ is analytic in a neighbourhood of $\infty$ and also analytic (and nonconstant) in a neighbourhood of $c_0$. For example, with $f(z) = z/(1+z)$ and $\sum_{j=0}^\infty z^j$ we have

$$ \eqalign{\dfrac{\sum_{j=0}^N z^j}{1 + \sum_{j=0}^N z^j} &= \dfrac{z^{N+1}-1}{z^{N+1}-z+2}\cr &\to 1 \ \text{for } |z| > 1\cr & \to \dfrac{1}{2-z} \ \text{for } |z| < 1\cr}$$

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This is very instructive. I'll take this in my mind to bed tonight and "let it work" inside. I'll come back to it later... – Gottfried Helms Dec 10 '12 at 23:00
My question arose when I re-read an older discussion of mine in the tetration-forum, where I didn't find an answer nor even a suitable direction for answer (for my level of understanding at that time) Here is the link to the discussion where I had posed this in context with iteration-series and landed at that example of my current question: – Gottfried Helms Dec 10 '12 at 23:11
Yes, I think things have now arrived, thank you for the explanations. Concerning "don't know what is your criterion" - it was just an unreflected transfer of my knowledge of infinite sums onto infinite products; where I should have remembered, that the inheritance of properties of divergence/convergence for the infinite products should be considered by analyzing the infinite sums of their logarithms (as was also stated in Knopp). I'm still not sure about the analoguous problem in powertowers to which I've linked in my previous comment- but lacking the analytic continuous iteration ... – Gottfried Helms Dec 11 '12 at 21:37
... the needed premises for analytical continuation are not yet established (and we do not yet know whether this shall be possible at all). So I think this answered my question here and also the other question in my tetration-forum link. – Gottfried Helms Dec 11 '12 at 21:38

Consider $f(z)=\dfrac{1}{1-z}$ as a function of one complex variable. Then $f$ is analytic near any $z\in\Bbb C$ except for $z=1,$ where it has a simple pole. The geometric series is the Taylor expansion of $f$ at $z=0,$ and because of the pole at $z=1,$ it has radius of convergence $1,$ meaning that $1+z+z^2+\cdots =\dfrac{1}{1-z}$ is valid for any complex number $|z|<1.$ For any other point $z_0$ in the complex plane, we can form a Taylor expansion $\sum_{n\ge 0}a_n(z-z_0)^n$ for $f$ with radius of convergence $R=|z_0-1|,$ but it will no longer be true that $a_n=1$ for all $n.$ The function $f$ is still well-defined and analytic at $z_0$ however.

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Could you please explain in which aspect it is relevant that "it will no longer be true that $a_n=1$ for all n"? I just don't get it... – Gottfried Helms Dec 11 '12 at 22:34
Dear @GottfriedHelms, sorry, it's a bit difficult to interpret what your original question asked. But, the statement $\sum x^n=\frac{1}{1-x}$ is not always true, and therefore one cannot simply replace $\frac{1}{1-x}$ by a geometric series in any formula. I thought this might be part of your question. – Andrew Dec 11 '12 at 22:45
E.g., in Robert Israel's post, he mentions that $f(\sum g_k)$ has analytic continuation to $\Bbb C\setminus\{1\},$ but the Taylor series of $g(z)=\frac{1}{1-z}$ near $z=2$ is not $\sum_{k=0}^\infty z^k,$ so we should definitely avoid, as he says, writing $e=\prod_k e^{-2^k}.$ – Andrew Dec 11 '12 at 22:54
Hi Andrew - it is not $\sum z^k$ - perhaps because of some recentering of the powerseries? – Gottfried Helms Dec 11 '12 at 23:16
Dear @GottfriedHelms, yes, exactly, we must centre it at $z=2$ (and if we instead write $e=\prod_k e^{(-a_k(z-2)^k)}$ for appropriate $a_k$'s, it is not so mysterious! We should have $a_0=-1,$ and all the other factors become $1$ when we substitute $z=2.$) – Andrew Dec 11 '12 at 23:23

You cannot analytically continue the sum in question beyond the unit disk, unless you omit $\{1\}$, in that your analytic continuation will not be valid at $x=1$. In your case, when you look at $e^{-1-2-4\ldots}$ you should in fact get 0 as you are raising an exponential to negative infinity.

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The series $\sum_{j=0}^\infty z^j$ converges for $|z|<1$, but it converges there to a function $1/(1-z)$ that has an analytic continuation to $\mathbb C \backslash \{1\}$. That's what he's referring to. – Robert Israel Dec 10 '12 at 22:18
@Robert Israel: thanks for clarification – Alex R. Dec 10 '12 at 22:21

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