How to get closed form from generating function?

Question

I have this generating function:

$$\frac{1}{2}\, \left( {\frac {1}{\sqrt {1-4\,z}}}-1 \right) \left( \,{ \frac {1-\sqrt {1-4\,z}}{2z}}-1 \right)$$

and I know that $\frac {1}{\sqrt {1-4\,z}}$ is the generating function for the sequence $\binom {2n} {n}$, and $\frac {1-\sqrt {1-4\,z}}{2z}$ is the generating function for the sequence $\frac {1}{n+1}\binom{2n} {n}$.

Now, I thought that I could substitute those in there, and where they multiply I'll use a summation like this:

$$\frac{1}{2}\left( 1-\frac{1}{n+1}\binom{2n} {n}-\binom{2n} {n} + \sum_{k=0}^n \frac{1}{k+1} \binom{2k}{k}\binom{2(n-k)}{n-k} \right)$$

Could this be right? It doesn't seem to work when I try in Maple. What else could I do?

I already know that the end sequence will be $\binom{2n-1}{n-2}$ if this can help...

Answer 1

Expand out the product, and look at what each term is the generating function for.

Answer 2

Expanding the product, and after some algebra: $$ g(z) = \frac{1}{2} \left(\frac{\sqrt{1-4 z}-1}{z}+\frac{1}{\sqrt{1-4 z}}+1\right) $$ Using generalized binomial theorem: $$ [z]^n g(z) = \frac{1}{2} \left( \delta_{n,0} + \binom{1/2}{n+1} (-4)^{n+1} + \binom{-1/2}{n} (-4)^n \right) $$ Using $$ \begin{eqnarray} \binom{-1/2}{n} - 4 \binom{1/2}{n+1} &=& \frac{\Gamma(1/2)}{n! \Gamma(1/2-n)} - \frac{\Gamma(3/2)}{(n+1)! \Gamma(1/2-n)} \\ &=& (-1)^n (n-1) \frac{\Gamma(1/2+n) \Gamma(n+1)}{\Gamma(1/2) (n+1)!n!} \\ &=& (-1)^n (n-1) \frac{(2n)!}{4^n (n+1)!n!} \\ &=& (-1)^n \frac{(2n-1)!}{2 \cdot 4^{n-1} (n+1)!(n-2)!} = 2 \cdot \frac{(-1)^n}{4^n} \cdot \binom{2n-1}{n+1} \end{eqnarray} $$ Hence, using Iverson's bracket: $$ [z]^n g(z) = \binom{2n-1}{n+1} \left[ n \ge 2 \right] $$

Answer 3

It’s not necessary to use the generalized binomial theorem and the gamma function. Let $$g(x)=\frac12\left(\frac1{\sqrt{1-4x}}-1\right) \left(\frac{1-\sqrt{1-4x}}{2x}-1\right)$$ and $u=\sqrt{1-4x}$. Then

$$\begin{align*} g(x)&= \frac12\left(\frac1u-1\right)\left(\frac{1-u}{2x}-1\right)\\ &=\frac12\left(\frac{1-u}{2xu}-\frac{1-u}{2x}-\frac1u+1\right)\\ &=\frac12\left(\frac1{2xu}-\frac1x+\frac{u}{2x}-\frac1u+1\right)\\ &=\frac12\left(\frac{1+u^2}{2xu}-\frac1x-\frac1u+1\right)\\ &=\frac12\left(\frac{1-2x}{xu}-\frac1x-\frac1u+1\right)\\ &=\frac12\left(\frac1{xu}-\frac1x-\frac3u+1\right)\\ &=\frac12\left(\frac1x\left(\frac1u-1\right)+1-\frac3u\right)\\ &=\frac12\left(\sum_{k\ge 1}\binom{2k}k x^{k-1}+1-3\sum_{k\ge 0}\binom{2k}k x^k\right)\\ &=\frac12\left(1+\sum_{k\ge 0}\left(\binom{2k+2}{k+1}-3\binom{2k}k\right)x^k\right)\\ &=\frac12\sum_{k\ge 2}\left(\binom{2k+2}{k+1}-3\binom{2k}k\right)x^k. \end{align*}$$

Now

$$\begin{align*} \binom{2k+2}{k+1}-3\binom{2k}k&=\binom{2k+1}k+\binom{2k+1}{k+1}-3\binom{2k}k\\ &=\binom{2k+1}k+\binom{2k}k+\binom{2k}{k-1}-3\binom{2k}k\\ &=\binom{2k+1}k+\binom{2k}{k+1}-2\binom{2k}k\\ &=\binom{2k}{k-1}+\binom{2k}k+\binom{2k}{k+1}-2\binom{2k}k\\ &=\binom{2k}{k-1}-\binom{2k}k+\binom{2k}{k+1}\\ &=\binom{2k}{k-1}-\binom{2k}k+\binom{2k-1}k+\binom{2k-1}{k+1}\\ &=\binom{2k}{k-1}-\binom{2k-1}{k-1}-\binom{2k-1}k+\binom{2k-1}k+\binom{2k-1}{k+1}\\ &=\binom{2k}{k-1}-\binom{2k-1}{k-1}+\binom{2k-1}{k+1}\\ &=\binom{2k-1}{k-2}+\binom{2k-1}{k+1}\\ &=2\binom{2k-1}{k-2}, \end{align*}$$

so $\displaystyle[x^k]g(x)=\binom{2k-1}{k-2}$, as desired.