How to evaluate the following sum?
$$ = \sum_{i \ge 0}^\infty \frac{i}{(a+i)^\underline{b}} $$
As an intermediate step in evaluating the sum, if we let:
$$ \Delta g(i) = \frac{1}{(a+i)^\underline{b}} $$
And assume $g(x) = 0, \forall x < 0$, is it valid to use finite calculus and say:
$$\begin{aligned}\Delta g(i) = \frac{1}{(a+i)^\underline{b}}\\ = (a+i)^\underline{-b}\end{aligned}$$
$$ g(i) = \frac{(a+i)^\underline{-b-1}}{-b}$$
Notice that $$\frac{n}{(n+a)^{\underline{b}}}=f(n-1)-f(n)$$ where $$f(n)=\frac{a+(b-1)n}{(b-1)(b-2)(n+a)^{\underline{b-1}}}$$ hence the given series is telescoping and is given by \begin{align} \sum_{n=0}^\infty\frac{n}{(n+a)^{\underline{b}}} &=\sum_{n=0}^\infty(f(n-1)-f(n))\\ &=f(-1)-\lim_{m\to\infty}f(m)\\ &=f(-1)\\ &=\frac1{(b-1)(b-2)(a-1)^{\underline{b-2}}}\\ \end{align}