Motivating example:
Consider the following $f(z)=\sum_{n\geq 0}A_n(z)$, and $g(z)=\sum_{n\geq 0} B_n(z)$.
Lets say $A_n(z)=\frac{(-1)^n}{(2n+1)^z}$ and $ B_n(z)=\frac{(-1)^{n}}{(n+1)^z}$
Then $f(z)=\beta(z)$ would be the Dirichlet beta and $g(z)=\eta(z)$ would be the Dirichlet eta function.
Let's say we KNOW what the values of $\eta(z)$ are after analytic continuation. $\eta(-1)=\frac{1}{4}$ and $\eta(0)=\frac{1}{2}$.
Now writing $\beta(-1)$ as its summation form would be nonsense, but if we were to, it would give us $$\beta(-1)=\sum_{n=0}^{\infty}(-1)^n(2n+1)=\sum_{n=0}^{\infty}[2(n+1)(-1)^n-(-1)^n]=2{\underbrace{\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^{-1}}}_{\eta(-1)}}-{\underbrace{\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^0}}_{\eta(0)}}=\frac{2}{4}-\frac{1}{2}=0$$
Which is the correct value of $\beta(-1)$.
I have encountered these quite a few times, where I am computing the analytic continuation of a function and I spot another function appearing where I know the value of it, the answer turns out to be correct when I use this technique. Therefore I would like to show the following :
Proposition
let $f(z)=\sum_{n\geq 0}A_n(z)$, and $g(z)=\sum_{n\geq 0} B_n(z)$.
If $g(z)$ has an analytic continuation on the neighbourhood of $z_i$ where $g(z_i)=w_i$.
If we have that $A_n(\alpha_0)=L(B_n(z_0),B_n(z_1),...,B_n(z_k))$ where $L$ is a linear function, then
$$f(\alpha_0)=L(g(z_0),...,g(z_k))$$
,
In our example previously. $L(x,y)=2x-y$
Proof
$$A_n(\alpha_0)=L(B_n(z_0),B_n(z_1),...,B_n(z_k))$$ $$\Rightarrow A_n(\alpha_0)-L(B_n(z_0),B_n(z_1),...,B_n(z_k))=0$$ \begin{equation} \Rightarrow \sum_{n=0}^{\infty}\left[A_n(\alpha_0)-L(B_n(z_0),B_n(z_1),...,B_n(z_k))\right]=0 \end{equation} $$f(\alpha_0)-L(g(z_0),...,g(z_k))\overset{(?)}{=}0$$
I am not sure if the third line to the fourth line is correct. $\alpha_0$ is a point whilst doing analytic continuation we should consider an open set. And notice that the linearity of the function $L$ depends on $\alpha_0$. $L$ would not be linear if $\alpha_0=\frac{1}{2}$ (say). As in our example, we can only expand $A_n(z)$ when $z$ is a non-positive integer, which we cannot do analytic continuation on (the set of non-positive integers is discrete).
Maybe we can define a function $C_n(\alpha_0,z_0,...,z_k)=A_n(\alpha_0)-L(B_n(z_0),B_n(z_1),...,B_n(z_k))$ and investigate $\sum C_n$? I am not sure how to proceede.
Any help is greatly appreciated.