I trying to figure out whether a simpler form of this series exists.
$$\sum_{i=0}^{n-2}\frac{L_{i+1}(-x)-L_{i}(-x)}{i+2}\left(\sum_{k=0}^{n-2-i} L_k(x)\right)$$
$L_n(x)$ is the $n$th Laguerre polynomial.
Wikipedia offers several useful recurrence relations in terms of the associated Laguerre polynomials. Two of which show that $$L_n^{(\alpha+1)}{x}=\sum_{i=0}^n{L_i^{(\alpha)}(x)}$$ and $$L_n^{(\alpha)}(x)=L_n^{(\alpha+1)}(x)-L_{n-1}^{(\alpha+1)}(x)$$ These allow us to simplify the expression a bit to $$\sum_{i=0}^{n-2}\frac{L_{i+1}^{(-1)}(-x)L_{n-2-i}^{(1)}(x)}{i+2}$$
Wikipedia also gives us the relation $$L_n^{(\alpha+\beta+1)}(x+y)=\sum_{i=0}^nL_{i}^{(\alpha)}(x)L_{n-i}^{(\beta)}(y)$$ Removing the denominator, our expression would completely reduce to $$\sum_{i=0}^{n-2}L_{i+1}^{(-1)}(-x)L_{n-2-i}^{(1)}(x)=L_{n-2}^{(1)}(0)=n-1$$ This is where I'm stuck. How might we evaluate it considering the denominator in the sum? Or is there some other way we can simplify the expression?
Let we set $N=n-2$. The generating function of Laguerre's polynomials is: $$ \sum_{n\geq 0}L_n(x)\,t^n = \frac{1}{1-t}\,\exp\left(-\frac{tx}{1-t}\right)\tag{1} $$ hence: $$ \sum_{n\geq 0}\left(\sum_{k=0}^n L_k(x)\right)\,t^n = \frac{1}{(1-t)^2}\,\exp\left(-\frac{tx}{1-t}\right)\tag{2} $$ $$ \sum_{n\geq 0}L_n(-x)\,t^n = \frac{1}{1-t}\,\exp\left(\frac{tx}{1-t}\right)\tag{3} $$ $$ \sum_{n\geq 0}\left(L_{n+1}(-x)-L_{n}(-x)\right)t^{n+1} = t\exp\left(\frac{tx}{1-t}\right)-t(1+x)\tag{4} $$ Now we may replace $t$ with $tu$ in $(4)$ and integrate both sides over $(0,1)$ with respect to $u$ to get: $$ \sum_{n\geq 0}\frac{L_{n+1}(-x)-L_n(-x)}{n+2}\,t^{n} \\= \frac{2-4t+2t^2-t^2x^2-t^2 x^3}{2t^2 x^2}+\frac{(1-t)(tx+t-1)}{t^2 x^2}\,\exp\left(\frac{tx}{1-t}\right)\tag{5}$$ and now: $$ \sum_{i=0}^{N}\frac{L_{i+1}(-x)-L_i(-x)}{i+2}\left(\sum_{k=0}^{N-i}L_k(x)\right)\tag{6a}$$ is the coefficient of $x^N$ in the product between the RHS of $(2)$ and the RHS of $(5)$.
It can be computed through $(1)$ with a bit of patience.