A sum with Laguerre polynomials

Question

I would like to know, if it is possible to simplify the following sum $$\sum_{n=1}^{\infty}\frac{L_{n}^{(0)}(x)}{n},$$ or, more generally, $$\sum_{n=0}^{\infty}\frac{(1-\alpha)_{n}}{(1+\alpha)_{n}}\frac{L_{n}^{(\alpha)}(x)}{n-\alpha},$$ for $x>0$ and $\alpha>-1$. Here $L_{n}^{(\alpha)}(x)$ is the (generalized) Laguerre polynomial: $$ L_{n}^{(\alpha)}(x)=\frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k}(\alpha+k+1)_{n-k}(-x)^{k}$$ and $(a)_{n}=a(a+1)\dots(a+n-1)$ is the Pochhammer symbol.

Answer 1

The series for the Laguerre polynomials converges absolutely as $L_n^{(\alpha)}(x)\sim n^{-\tfrac{1}{4}}$ as $n\to\infty$ (see their asymptotic approximations DLMF). Defining \begin{equation} S(z)=\sum_{n=1}^\infty \frac{L_n(x)}{n}z^n \end{equation} one has \begin{align} S'(z)&=\sum_{n=1}^\infty L_n(x)z^{n-1}\\ &=\frac{1}{z}\sum_{n=1}^\infty L_n(x)z^{n}\\ &=\frac{1}{z}\left[(1-z)^{-1}\exp\left(\frac{xz}{z-1}\right)-1\right] \end{align} where we use the generating function for the Laguerre polynomials \begin{equation} (1-z)^{-\alpha-1}\exp\left(\frac{xz}{z-1}\right)=\sum_{n=0}^{\infty}L^{(\alpha% )}_{n}\left(x\right)z^{n} \end{equation} with $\alpha=0$ and $L_0(x)=1$. Then the summation is \begin{equation} S(1)=\int_0^1 \frac{dz}{z}\left[(1-z)^{-1}\exp\left(\frac{xz}{z-1}\right)-1\right] \end{equation} with $u=z/(1-z)$, one obtains \begin{align} S(1)&=\int_0^\infty \frac{du}{u(1+u)}\left[(1+u)e^{-ux}-1\right]\\ &=lim_{\epsilon\to0^+}[E_1(\epsilon x)-ln(\epsilon)-\ln(1+\epsilon)]\\ &=-\gamma-\ln x \end{align}