Let $s$ with $Re(s)> 1$ and let $d>1$ be an integer. Consider the series $$ f(s) = \sum_{k=0}^\infty \frac{\binom{k + d -1}{k}}{(1+k)^{ds}} $$
I would like to show that the residue of $f$ at $s = 1$, i.e. $\lim_{s \to 1^+} (s-1)f(s)$, is $1/ d!$. This computation arises when trying to compute the Dixmier trace of the operator $\Delta^{-d}$ of multiplication by $(n_1 + \cdots + n_d)^{-d}$ in $L^2(\mathbb{Z}_+^d)$.
Using $b^{-a}=\frac{1}{\Gamma(a)}\int_0^\infty x^{a-1}e^{-bx}~dx$ and $\sum_{k=0}^\infty\binom{k+d-1}{k}z^k=(1-z)^{-d}$, you get $$f(s)=\frac{1}{\Gamma(ds)}\int_0^\infty\frac{x^{ds-1}e^{-x}}{(1-e^{-x})^d}~dx,$$ much like the "textbook" case $d=1$, when $f(s)=\zeta(s)$; now we split $\int_0^\infty=\int_0^1+\int_1^\infty$ (say), neglect the $\int_1^\infty$ (as it is regular at $s=1$), and expand the integrand for $\int_0^1$ into power series (for $s>1$): $$\int_0^1\frac{x^{ds-1}e^{-x}}{(1-e^{-x})^d}~dx=\int_0^1(x^{d(s-1)-1}+\ldots)~dx=\frac{1}{d(s-1)}+\mathcal{O}(1),\quad s\to1^+.$$ Together with $\Gamma(d)=(d-1)!$, this gives the desired $\lim\limits_{s\to1^+}(s-1)f(s)=1/d!$.
