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

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