Put
$$ S_n(\xi, x)=\sum_{k=0}^{+\infty} \frac{L_k^n(x)}{\xi-k}, x \in C, \xi \in C \backslash Z_{+} $$ Now, to compute the sum $S_n(\xi, x)$ we begin by writing it in terms of the Laguerre polynomial of order zero thanks to the identity $$ L_k^n(x)=(-1)^n \frac{d^n}{d x^n}\left(L_{n+k}(x)\right) $$ Precisely, we have $$ \begin{aligned} S_n(\xi, x) & =(-1)^n\left(\frac{d}{d x}\right)^n\left(\sum_{k=0}^{\infty} \frac{L_{k+n}(x)}{\xi-k}\right) \\ & =(-1)^{n+1}\left(\frac{d}{d x}\right)^n\left(\sum_{j=0}^{+\infty} \frac{L_j(x)}{j-(n+\xi)}\right), \end{aligned} $$
which must be interpreted in the distributional sense.
I have read this in a paper and I don't understant why the authors have changed the derivative with sum see the following $S_n(\xi, x)=(-1)^{n+1}\left(\frac{d}{d x}\right)^n\left(\sum_{j=0}^{+\infty} \frac{L_j(x)}{j-(n+\xi)}\right),$