We observe the double sum:
$$\sum_{n=1}^{\infty}\sum_{k=0}^{m}(-1)^k{m\choose k}{n-ka\over (n+m-k)^m}=f(m,a),m\ge2\tag1$$ $a$ is not restricted
We are trying to determine the closed form for $(1)$
Let expanded $(1)$ for $m=2,3$ and $4$
$$\sum_{n=1}^{\infty}{n\over (n+2)^2}-2\cdot{n-a\over (n+2)^2}+{n-2a\over n^2}=f(2,a)$$
$$\sum_{n=1}^{\infty}{n\over (n+3)^3}-3\cdot{n-a\over (n+2)^3}+3\cdot{n-2a\over (n+2)^3}-{n-3a\over n^3}=f(3,a)$$
$$\sum_{n=1}^{\infty}{n\over (n+4)^4}-4\cdot{n-a\over (n+3)^4}+6\cdot{n-2a\over (n+2)^4}-4\cdot{n-3a\over (n+1)^4}+{n-4a\over n^4}=f(4,a)$$
We was able to determine the closed form for
$$f(2,a)=1-2a$$ $$f(3,a)={7\over8}(3a-1)$$ $$f(4,a)={575\over 648}(1-4a)$$
By the look of it, we can assume the general closed form $(1)$ might take the form of $$f(m,a)=(-1)^m(1-ma)g(m)$$
How can we find the general closed form for $(1)$?
