Help find closed form: $\sum_{n=1}^{\infty}\sum_{k=0}^{m}(-1)^k{m\choose k}{n-ka\over (n+m-k)^m}$

Question

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 $(1)$?

Answer 1

We have $$ f(m,a)=(-1)^m(ma-1)g(m) $$ with $$ g(m)=\sum_{j=1}^{m-1}\frac{(-1)^j}{j^m}\binom{m-2}{j-1}. $$ For this take the initial expression of $f(m,a)$, exchange the sums, and use the following formula $$ (1)\qquad\qquad\sum_{n=1}^{\infty}\frac{n-ka}{(n+m-k)^m}=\zeta(m-1)+(k(1-a)-m)\zeta(m)-\sum_{j=1}^{m-k}\frac{1}{j^{m-1}} \qquad-(k(1-a)-m)\sum_{j=1}^{m-k}\frac{1}{j^{m}}, $$ where $\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$. Then use that $\sum_{k=0}^{m}(-1)^k k\binom{m}{k}=0=\sum_{k=0}^{m}(-1)^k \binom{m}{k}$, and do some straightforward computations. Finally use the following formulas $$ (2)\qquad\qquad\qquad \sum_{k=0}^{j}(-1)^k k\binom{m}{k} =(-1)^j m\binom{m-2}{j-1} $$ and $$ (3)\qquad\qquad\qquad \sum_{k=0}^{j}(-1)^k \binom{m}{k} =(-1)^j \binom{m-1}{j}, $$ to obtain the result.

Note that for $j=m$ the formulas (2) and (3) yield $\sum_{k=0}^{m}(-1)^k k\binom{m}{k}=0=\sum_{k=0}^{m}(-1)^k \binom{m}{k}$.