Generating function of binomial coefficients

Question

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.

Answer 1

Note: The cited formula (8) is not correct. There is a typo since we have \begin{align*} \sum_{n=0}^{\infty}\binom{n}{k}y^n = \frac{y^{\color{blue}{k}}}{(1-y)^{\color{blue}{k+1}}}\tag{1} \end{align*} We therefore expect \begin{align*} \frac{1}{x}\sum_{L=0}^{\infty}\binom{L+1}{2}x^{L+1} =\frac{1}{x}\sum_{L=1}^{\infty}\binom{L}{2}x^L = \frac{x^{\color{blue}{1}}}{(1-x)^{\color{blue}{3}}}\tag{2} \end{align*}

Your derivation is fine up to
\begin{align*} \frac{d}{dx}& \frac{2x-x^2}{(1-x)^2} - \frac{1}{(1-x)^2} \\ &=\frac{1}{2}\left((2-2x)(1-x)^{-2}+(2x-x^2)(+2)(1-x)^{-3}\right)-\frac{1}{(1-x)^2}\\ &=\left(\frac{1}{1-x}+\frac{2x-x^2}{(1-x)^3}\right)-\frac{1}{(1-x)^2}\\ &=\frac{1}{(1-x)^3}-\frac{1}{(1-x)^2}\\ &\,\,\color{blue}{=\frac{x}{(1-x)^3}} \end{align*}

in accordance with (2).

Hint: With respect to (1) we have \begin{align*} \frac{y^k}{(1-y)^{k+1}}&=y^k\sum_{n=0}^\infty \binom{-(k+1)}{n}(-y)^n\tag{3}\\ &=\sum_{n=0}^\infty \binom{k+n}{n}y^{n+k}\tag{4}\\ &=\sum_{n=k}^\infty \binom{n}{n-k}y^n=\sum_{n=k}^\infty\binom{n}{k}y^n\tag{5}\\ &=\sum_{n=0}^\infty\binom{n}{k}y^n\tag{6} \end{align*}

Comment:

• In (3) we use the binomial series expansion.

• In (4) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (5) we shift the index to start with $n=k$.

• In (6) we use $\binom{p}{q}=0$ if $q>p$.

Answer 2

In my humble opinion, a simpler way would to consider

$$S=\sum_{L=0}^{\infty}\frac{1}{2}L(L+1)x^L\implies T=2S=\sum_{L=0}^{\infty}L(L+1)x^L$$ $$T=\sum_{L=0}^{\infty}L(L-1+2)x^L=\sum_{L=0}^{\infty}L(L-1)x^L+2\sum_{L=0}^{\infty}Lx^L$$

$$T=x^2\sum_{L=0}^{\infty}L(L-1)x^{L-2}+2x\sum_{L=0}^{\infty}Lx^{L-1}$$ $$T=x^2 \left(\sum_{L=0}^{\infty}x^{L} \right)''+2x \left( \sum_{L=0}^{\infty}x^{L}\right)'$$