we have the summation : $$\underset{n\neq m}{\sum_{n=1}^{\infty}}\frac{n^{j-1}}{\left(n-m\right)^{j+1}}$$ where $j,m$ are positive integers . By partial fraction expansion, we have: $$\underset{n\neq m}{\sum_{n=1}^{\infty}}\frac{n^{j-1}}{\left(n-m\right)^{j+1}}=\sum_{i=0}^{j-1}\binom{j-1}{i}\underset{n\neq m}{\sum_{n=1}^{\infty}}\frac{m^{j-1-i}}{(n-m)^{j-i+1}}$$ We use the series expansion of the polygamma function to write the rightmost summation in terms of the limit: $$\underset{n\neq m}{\sum_{n=1}^{\infty}}\frac{1}{(n-m)^{j-i+1}}=\lim_{z\rightarrow m}(-1)^{j-i+1}\frac{\psi^{(j-i)}(1-z)}{(j-i)!}-\frac{1}{(m-z)^{j-i+1}}$$ or, in terms of the Hurwitz Zeta Function: $$\lim_{z\rightarrow m}\zeta(j-i+1,1-z)-\frac{1}{(m-z)^{j-i+1}}$$ 1)Is there a closed form for limit!?
2)Is there another way we can write our summation in closed form !?
$$\begin{align} \sum_{n=1,n\neq m}^\infty\frac{n^{j-1}}{(n-m)^{j+1}} &=\sum_{n=1}^{m-1}\frac{n^{j-1}}{(n-m)^{j+1}}+\sum_{n=m+1}^\infty\frac{n^{j-1}}{(n-m)^{j+1}}\\ \\&\phantom{=}\text{This is how I re-index; literally replace every $n$.} \\ &=\sum_{n=1}^{m-1}\frac{n^{j-1}}{(n-m)^{j+1}}+\sum_{n+m=m+1}^{n+m=\infty}\frac{(n+m)^{j-1}}{((n+m)-m)^{j+1}}\\ &=\sum_{n=1}^{m-1}\frac{n^{j-1}}{(n-m)^{j+1}}+\sum_{n=1}^{\infty}\frac{(n+m)^{j-1}}{n^{j+1}}\\ \\&\phantom{=}\text{Use the Binomial Theorem.} \\ &=\sum_{n=1}^{m-1}\frac{n^{j-1}}{(n-m)^{j+1}}+\sum_{n=1}^{\infty}\frac{1}{n^{j+1}}\sum_{k=0}^{j-1}\binom{j-1}{k}n^{j-1-k}m^k\\ &=\sum_{n=1}^{m-1}\frac{n^{j-1}}{(n-m)^{j+1}}+\sum_{n=1}^{\infty}\sum_{k=0}^{j-1}\binom{j-1}{k}n^{-2-k}m^k\\ &=\sum_{n=1}^{m-1}\frac{n^{j-1}}{(n-m)^{j+1}}+\sum_{k=0}^{j-1}\binom{j-1}{k}m^k\sum_{n=1}^{\infty}n^{-2-k}\\ &=\sum_{n=1}^{m-1}\frac{n^{j-1}}{(n-m)^{j+1}}+\sum_{k=0}^{j-1}\binom{j-1}{k}m^k\zeta(k+2)\\ \end{align}$$