Are all analytic continuations consistent with $1-1+1-1+\dots = \frac{1}{2}$?

Question

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 $g(z_0)=\frac{1}{2}$?

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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 $a_n(z)=z^n,z_0=-1$.

Answer 1

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:

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

Answer 2

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.