In Abel's famous article "Studies on Elliptic Functions", most of which I've understood, there is a formula that confused me. It is like this: $$f(x)-f(x+1)+f(x+2)-\cdots = 0.5f(x)+Af'(x)+Bf''(x)+\cdots$$
Abel said this is a known equation, but I don't know how to deduce it. For $A, B$, he just said "where $A, B, ...$ are numbers", nothing else.
If I use Taylor expansion for the left part on $x=x$, we can get:
$$\begin{align} f(x)&=f(x) \\[4pt] f(x+1)&=f(x)+f'(x)\cdot1+\frac{1}{2!}f''(x)\cdot 1^2+\frac{1}{3!}f'''(x)\cdot1^3+\cdots\\[4pt] f(x+2)&=f(x)+f'(x)\cdot2+\frac{1}{2!}f''(x)\cdot 2^2+\frac{1}{3!}f'''(x)\cdot2^3+\cdots\\[4pt] \cdots&=\cdots\\[4pt] f(x+k)&=f(x)+f'(x)\cdot k+\frac{1}{2!}f''(x)\cdot k^2+\frac{1}{3!}f'''(x)\cdot k^3+\cdots \end{align}$$ and $$\begin{align} f(x)-f(x+1)+f(x+2)-f(x+3)+ \cdots &= (1-1+1-1+\cdots)f(x) \\[4pt] &+(-1+2-3+\cdots)f'(x) \\[4pt] &+\frac12(-1^2+2^2-3^2+\cdots)f''(x)\\[4pt] &+\frac{1}{6}(-1^3+2^3-3^3+\cdots)f'''(x)\\[4pt] &+\cdots \end{align}$$
So refer to Abel's equation, we can see: $$\begin{align} 1-1+1-1+\cdots&=0.5 \\ -1+2-3+\cdots&=A \\ -1^2+2^2-3^2+\cdots&=B \end{align}$$ This is strange because the left series are divergent.
so I just want to know how this equation is deduced.
In the original article, the function $f(x)$ is $$f(m,x)=\frac{2\alpha}{\alpha^2-((m+\frac{1}{2})w+(x+\frac{1}{2})w'\sqrt{-1})^2}$$ $\alpha,m,w,w'$ can be seen as real constants.
This question is important for Abel's deduction and to have a good understanding of elliptic functions, as I haven't seen anything similar in the textbook, this technique may be used in other research. In fact, in his article, there are many print mistakes, it is $Bf'''(x)$ in the original article.
However, by the meaning of Abel, it seems that the equation doesn't depend on the concrete form of f(x). Because he used this equation as a known equation as he said.
If you are still not clear, https://www.maa.org/sites/default/files/images/upload_library/1/abeltranslation.pdf
Download this paper and at the end of page 64, you can find it, this part is self-contained and doesn't depend too much on the context.
After some investigation about the history and under the help of Wiki, I found a good explanation of this question at last. By the famous formula: $$1-2^n+3^n-4^n+... = \frac{2^{n+1}-1}{n+1}B_{n+1}$$ We can use it to calculate and get your desired result.