Strange summation? $\sum _{k=-n}^{n+1} \frac{(-1)^k}{x-k}$

Question

I'm mainly concerned with the bounds of summation here. I've never personally seen such a summation before, but I came across this summation in "Special Functions" by Andrews Askey Roy on page 11. Can someone explain to me or tell me laconically the name of this type of summation? I can do the remainder of the research It reminds me of proving a series of a sequence of functions converging uniformly (via Cauchy Condition), but this (of course) involves different bounds of summation. I imagine this is a property of summation; the same type of property used in Cauchy's Condition for a uniformly convergent series of sequence of functions. To be clear: $$\bigg|\sum_{k=0}^{n} f_{k}(x)-\sum_{k=0}^{m}f_{k}(x)\bigg|$$ $$=\bigg|\sum_{k=m+1}^{n}f_{k}(x)\bigg|$$ But, with m=-n+1 $$=\bigg|\sum_{k=-n}^{n}f_{k}(x)\bigg|$$ Am I correct? Isn't this similar? This was half(probably not half) of a proof that Roy was doing. He was attempting to show that $\frac{\pi}{\sin (\pi x)}=\frac{1}{x}+2x\sum_{n=1}^{\infty}\frac{(-1)^n}{x^2-n^2}=\lim_{n\to\infty}\sum_{k=-n}^{n}\frac{(-1)^k}{x-k}$. He uses the the integral$$\int_{0}^{\infty} \frac{t^{x-1}}{1+t}dt=\int_{0}^{1}\frac{t^{x-1}}{1+t} dt +\int_{1}^{\infty}\frac{t^{x-1}}{1+t}dt$$ In fact, I was able to get myself that the right hand integral (the second integral on the right-hand side) is $$\lim_{s\to\infty}\sum_{k=0}^{\infty}\frac{(-1)^{k}s^{x+k}}{x+k}-\sum_{k=0}^{\infty}\frac{(-1)^k}{x+k}$$

Answer 1

Hint
Write $$ T_n := \sum_{k=-n}^{n+1} \frac{(-1)^k}{x-k} $$ then for $0 < m < n$ we have $$ T_n-T_m = \sum_{k=-n}^{-m-1} \frac{(-1)^k}{x-k} + \sum_{k=m+2}^{n+1} \frac{(-1)^k}{x-k} $$ Then use the usual Cauchy condition for sequences on the sequence $\{T_n\}$.