<s>[INST] Generating function of binomial coefficients We want to evaluate the sum $$\sum_{L=0}^{\infty}\frac{1}{2}L(L+1)x^L$$
From this set of notes (page 2, equation 8) we find the formula $$\sum_{n=0}^{\infty}\binom{n}{k}y^n = \frac{y^n}{(1-y)^{n+1}}$$ which suggests that I can do $$\frac{1}{x}\sum_{L=0}^{\infty}\binom{L+1}{2}x^{L+1} = \frac{x^L}{(1-x)^{L+2}}\tag{1}\label{eqn1}$$ since $$\frac{1}{2}L(L+1) = \frac{(L+1)!}{2!((L+1)-2)!}$$
But if I do 
\begin{aligned}
\sum_{L=0}^{\infty}\frac{1}{2}L(L+1)x^L & = \sum_{L=0}^{\infty}\frac{L}{2}\frac{d}{dx}x^{L+1} \\ 
& = \sum_{L=0}^{\infty} \frac{1}{2} \frac{d}{dx} \left[(L+2)x^{L+1} - 2x^{L+1}\right] \\
& = \frac{1}{2} \frac{d^2}{dx^2} \sum_{L=0}^{\infty} x^{L+2} - \frac{d}{dx} \sum_{L=0}^{\infty} x^{L+1} \\
& = \frac{1}{2} \frac{d^2}{dx^2} \frac{x^2}{1-x} - \frac{d}{dx} \frac{x}{1-x} \\
& = \frac{1}{2} \frac{d}{dx} \left[2x(1-x)^{-1} + x^2(1-x)^{-2}\right] - \left[(1-x)^{-1} + x(1-x)^{-2}\right] \\
& = \frac{1}{2} \frac{d}{dx} \frac{2x-x^2}{(1-x)^2} - \frac{1}{(1-x)^2} \\
& = \frac{1}{2} \frac{d}{dx} \left[\frac{x}{(1-x)^2} + \frac{x}{1-x} \right] - \frac{1}{(1-x)^2} \\
& = \frac{1}{2} \left[(1-x)^{-2} + 2x(1-x)^{-3} \right] \\
& = \frac{1+x}{2(1-x)^3}
\end{aligned}
which is nowhere close to (1). Where did I go wrong?
All help is welcome.
 [/INST] In my humble opinion, a simpler way would to consider
$$S=\sum_{L=0}^{\infty}\frac{1}{2}L(L+1)x^L\implies T=2S=\sum_{L=0}^{\infty}L(L+1)x^L$$
$$T=\sum_{L=0}^{\infty}L(L-1+2)x^L=\sum_{L=0}^{\infty}L(L-1)x^L+2\sum_{L=0}^{\infty}Lx^L$$
$$T=x^2\sum_{L=0}^{\infty}L(L-1)x^{L-2}+2x\sum_{L=0}^{\infty}Lx^{L-1}$$
$$T=x^2 \left(\sum_{L=0}^{\infty}x^{L} \right)''+2x \left( \sum_{L=0}^{\infty}x^{L}\right)'$$
</s>