A sum with Laguerre polynomials 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.
 A: 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}
