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Background:

This is a follow-up to this question, which is the same question except for a different series (and I suspect the answer may be different for this series—more on that later). In short, I am curious about whether the idea of associating divergent sums with analytic continuations could be used to rigorously define a summation method for divergent series.

Question:

Let $A$ and $B$ be open subsets of $\mathbb C$ with $A\subset B$ and $\{a_n\}_{n\in\mathbb N_0}$ be a family of analytic functions $a_n:B\to\mathbb C,z\mapsto a_n(z)$ such that:

  1. $a_n(z_0)=(-1)^n$ for all $n\in\mathbb N_0$ and for some $z_0\in B$,
  2. $f(z):=\sum_{n=0}^\infty a_n(z)$ is well defined and analytic on $A$.

Let $g$ be the analytic continuation of $f$ to the rest of $\mathbb C$. Must $g(z_0)=\frac{1}{2}$?

The linked question above asks the same thing for $1+2+3+\dots=-\frac{1}{12}$, and a comment links to another question that provides a counter-example. So the answer is “no” in that case. But I suspect it may be “yes” in this case because $1-1+1-\dots=\frac{1}{2}$ is better behaved in that it is Cesaro mean convergent.

Additionally there are already two examples of such analytic continuations that agree with this sum in the sense described above: (1) The Dirichlet eta function, with $a_n(z)=\frac{(-1)^n}{n^z},z_0=0$, and (2) The geometric series function, with $a_n(z)=z^n,z_0=-1$.

WillG
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  • Well obviously you need more restriction on $a_n$, else one can simply define a piecewise $a_n$ s.t. $a_n(z_0)$ doesn't relate to $f$ at all. – Simply Beautiful Art Nov 27 '19 at 17:05
  • Good point, I’ve edited the question to make the $a_n$ functions analytic. I think now it makes sense. – WillG Nov 27 '19 at 19:21
  • Do you know the essay on "analytic continuation" on kmath https://www.mathpages.com/home/kmath649/kmath649.htm Due to this article I doubt whether different versions/the plural of "analytic continuation" exist at all. It seems to me it is a fixed technical term with the exact meaning of a very specific structure... – Gottfried Helms Nov 30 '19 at 15:17
  • If we begin with a divergent sum and want to asssign it to a value, I am pretty sure that we always have tons of possibilities and can in principle achieve any desired value. The regularizations are just one way to do that. And we should avoid to write it like an equation making sense. It is only an assignment, not an equality. – Peter Jan 03 '21 at 15:48

2 Answers2

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Consider the function:

$$f(s)=\sum_{n=1}^\infty\left(\frac1{(2n-1)^s}-\frac{1+\frac sn}{(2n)^s}\right)$$

Then it is obvious that we have

$$f(s)=\eta(s)-\frac s{2^s}\zeta(s+1)$$

which, as $s\to0$, gives:

$$\sum_{n=1}^\infty(-1)^{n+1}\sim-\frac12$$

Simply Beautiful Art
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  • I’m confused about the LHS of your last line. With your $a_n(s)$, I’m getting $a_n(0)=\frac{1}{(2n-1)^0}-\frac{1+0}{(2n)^0}=\frac{1}{1}-\frac{1}{1}=0.$ – WillG Nov 27 '19 at 19:31
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    Ok now I see what you’re getting at, because each $a_n(0)=1-1$. But this technically doesn’t meet the criteria for my $a_n$ functions, and I think that’s important because we can’t a priori assume this sort of “associative property” holds when summing divergent series. – WillG Nov 27 '19 at 19:34
  • $a_{2n-1}(s)=\frac1{(2n-1)^s}$ and $a_{2n}(s)=-\frac{1+\frac sn}{(2n)^s}$. – Simply Beautiful Art Nov 27 '19 at 19:35
  • Ah, I see. That does the trick. So the answer to my question is still “no.” – WillG Nov 27 '19 at 19:38
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    @WillG I think there's a flaw in this answer. $\sum_{n=1}^\infty (-1)^{n+1} = \sum_{n=0}^{\infty - 1} (-1)^{n+2} = \sum_{n=0}^{\infty - 1} (-1)^n \neq \sum_{n=0}^\infty (-1)^n$. There's a "missing term at infinity", so to speak. – user76284 Jan 21 '23 at 03:37
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Analytic continuation is a very fragile method that very heavily depends on the function you choose to continue.

Unlike those other major methods that usually give the same result, you often can arrive at what results you want with analytic continuation.

Possibly, it can be used to more rigorously define summation of divergent series but one should add additional conditions on the behavior of the functions they use to analytically continue. So far, I am unaware of any such work.

On the other hand, it seems, analytic continuation of Zeta function and other Dirichlet series works quite safe.

Addendum regarding notation

When you write (for instance) $1+1+1+\dots$, it is unclear what series do you mean, it can de different ones. For instance, it can be

$\sum_{k=0}^\infty 1 =\frac12$ or $\sum_{k=1}^\infty 1 =-\frac12$

You can test the both in Mathematica computer algebra system with Dirichlet regularization:

Sum[1,{x,0,Infinity},Regularization->Dirichlet]
Sum[1,{x,1,Infinity},Regularization->Dirichlet]

In your case you use alternating signs, so the series is stable-method-summable, (for instance, with Cesaro or Abel), so it sums to $\frac12$, but again, please use exact notation.

Anixx
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