1-1+1-1+... = k+1/2?

Question

Let $$F(s) = \sum_{n=0}^{\infty} f_{n}(s)$$ be a complex analytical function defined by a series (not necessarily a power series) that absolutely converges on an open set $s \in U$. Assume that the function $F$ can be analytically continued to an $s_0$ such that it holds $f_n(s_0) = (-1)^n$. Prove or refute that $$F(s_0) = k + 1/2$$ for some $k \in \mathbb{Z}$

Some time ago I was reading Terence Tao's blog about analytic continuation and he said the formal value of $1 - \frac{1}{2^s} + \frac{1}{3^s} - \cdots$ at $s=0$ is $\frac{1}{2}$. I knew from before that the geometric series $1 - s + s^2 - \cdots = \frac{1}{1+s}$ can be analytically continued to $s=1$ to give also $\frac{1}{2}$. Two seemingly unrelated series give the same value for Grandi's series and I was wondering if this was true in general. The answer was NO because there are symmetry transformations that map the Grandi's series to another ones that look identical, but such a transformation applied on the original, convergent series could alter the function and its continuation. For illustration I refer you to the accepted answer in the linked question right above.

I can recognize two symmetry transformations that leave Grandi's series looking the same:

1. Interchanging pairs of terms with both even or both odd indices in the sum. This transformation leaves both the geometric series and Terry's series unchanged because of their absolute convergence within their respective domains.
2. Permuting all pairs of terms of indices $2n$ and $2n+1$, and then adding 1 at the beginning. This has the effect of adding one to the series. The inverse can be applied to subtract 1.

If I didn't miss any symmetry, would then the statement above be true?

Answer 1

The result is not true as for example the Koebe function $K(z)=\frac{z}{(1-z)^2}=\sum_{n \ge 1} nz^n$ can be continued at $-1$ and $K(-1)=-1/4$ while choosing a sequence of natural numbers $g$ st $g$ preserves parity and the cardinality of the set $g^{-1}(m)$ is $m$ for each $m \ge 1$ (eg by induction starting with $g(0)=g(2)=2, g(1)=1, g(3)=g(5)=g(7)=3$ etc) gives $K=\sum_{n \ge 0}z^{g(n)}$, so choosing $f_n(z)=z^{g(n)}, s_0=-1$ gives a counterxample

Of course, this can be easily generalized to any $f(z)=\sum a_nz^n, a_n \in \mathbb N$ that converges on a small disc around the origin and has an analytic continuation on a domain containing the original disc and $-1$ (by choosing $g$ as above preserving parity and st the cardinality of the set $g^{-1}(m)$ is $a_m$ which again can be easily done by induction and putting $f=\sum z^{g(n)}$), so in particular all the derivatives of $\frac{1}{1-z}$ work giving functions that take $f(-1)=2^{-k}$ and then one can sum finitely such so get all rational numbers as possible $f(-1)$ for functions satisfying the OP conditions; would expect variations on this theme to give also irrational possible values