In the paper A Probabilistic Algorithm for k-SAT Based on Limited Local Search and Restart, by Uwe Schöning, I fail to understand an identity used in a proof: $$\sum_{i=0}^{\infty}\binom{2i+j}{i}z^i=\frac{B_2(z)^j}{\sqrt{1-4z}}$$
for $z=q(1-q)$, with
$$B_2(z)=\sum_{i=0}^{\infty}\binom{2i+1}{i}\frac1{2i+1}z^i=\frac{1-\sqrt{1-4z}}{2z}$$
and
$$B_2(z)^r=\sum_{i=0}^{\infty}\binom{2i+r}{i}\frac{r}{2i+r}z^i$$
It would be great if someone could explain this to me.
This is an interesting question which is very much related to a much more general question Closed form solutions for a family of hypergeometric sums. . Seeing that there are other users in here who are too concerned with summing up hypergeometric sums gives me the motivation to provide an answer. Hopefully this will boost other users to tackle my question as well. First of all let us notice that: \begin{equation} \binom{2 i+j}{i} = \binom{2 i}{i} \cdot \frac{(2i+1)^{(j)}}{(i+1)^{(j)}} \end{equation} for $j=1,2,\cdots$. Here $x^{(j)}:= x(x+1)\cdot \dots \cdot (x+j-1)$ is the upper Pochhammer symbol.In other words the binomial factor is a product of a simple binomila factor and a a ratio of two Pochhammer symbols.
On the other hand from the definition of the factorial we have: \begin{equation} \binom{2 i}{i} = \frac{4^i i! (i-1/2)!}{(-1/2)!} \cdot \frac{1}{i! i!}=4^i \binom{i-1/2}{i} = (-4)^i \binom{-1/2}{i} \end{equation} and therefore we get immediately that \begin{equation} \sum\limits_{i=0}^\infty \binom{2 i}{i} z^i = \left(1-4 z \right)^{-\frac{1}{2}} \end{equation} which is the result for $j=0$.
Now let us try to derive the result by induction in $j$. The tactic is to decompose the ratio of Pochhammer symbols into partial fractions in the variable $i$ and then use integration tricks to find closed form solutions for the sum.
Take $j=1$. Then: \begin{eqnarray} \frac{2 i+1}{i+1} &=& \left(2- \frac{1}{i+1}\right)\\ \Longrightarrow &&\\ rhs &=& 2\left(1-4 z\right)^{-\frac{1}{2}} - \underbrace{\int\limits_0^1 \left(1-4 z \theta \right)^{-\frac{1}{2}} d \theta}_{\frac{2}{1+\sqrt{1-4 z}}}= \frac{1}{\sqrt{1-4 z}} \left(\frac{1-\sqrt{1-4 z}}{2 z}\right) \end{eqnarray} as it should be. Take $j=2$. Then: \begin{eqnarray} \frac{(2 i+1)^{(2)}}{(i+1)^{(2)}} &=& \left(4- \frac{6}{i+2}\right)\\ \Longrightarrow &&\\ rhs&=& 4\left(1-4 z\right)^{-\frac{1}{2}}-6\underbrace{\int\limits_0^1 \theta \left(1-4 z \theta \right)^{-\frac{1}{2}} d \theta}_{\frac{1-\sqrt{1-4 z}(1+2 z)}{12 z^2}} = \frac{1}{\sqrt{1-4 z}} \left(\frac{1-\sqrt{1-4 z}}{2 z}\right)^2 \end{eqnarray} as it should be. Now let us assume that $j$ is an odd number. Then the partial fraction decomposition, in the variable $i$, we are after reads: \begin{eqnarray} \frac{(2i+1)^{(2j+1)}}{(i+1)^{(2j+1)}} &=& 2^{2j+1} \left(1+\sum\limits_{\xi=0}^j (\xi+\frac{1}{2}) \binom{-\frac{3}{2}-\xi}{j} \binom{j}{\xi} (-1)^{1+\xi} \cdot \frac{1}{i+j+\xi+1}\right)\\ \end{eqnarray} Therefore we have: \begin{eqnarray} &&rhs=\\ && 2^{2j+1} \left((1-4 z)^{-\frac{1}{2}}+\sum\limits_{\xi=0}^j (\xi+\frac{1}{2}) \binom{-\frac{3}{2}-\xi}{j} \binom{j}{\xi} (-1)^{1+\xi} \cdot \int\limits_0^1 \theta^{j+\xi} \left(1-4 z \theta \right)^{-\frac{1}{2}} d \theta\right)=\\ && 2^{2j+1} \left((1-4 z)^{-\frac{1}{2}}+\sum\limits_{\xi=0}^j (\xi+\frac{1}{2}) \binom{-\frac{3}{2}-\xi}{j} \binom{j}{\xi} (-1)^{1+\xi} \cdot \frac{2}{(4 z)^{j+\xi+1}} \int\limits_{\sqrt{1-4 z}}^1 (1-u^2)^{j+\xi} du\right)=\\ && 2^{2j+1} \left((1-4 z)^{-\frac{1}{2}}+\sum\limits_{\xi=0}^j \sum\limits_{p=0}^{\xi+j} (\xi+\frac{1}{2}) \binom{-\frac{3}{2}-\xi}{j} \binom{j}{\xi} (-1)^{1+\xi} \cdot \right.\\ &&\left. \frac{2}{(4 z)^{j+\xi+1}} \binom{j+\xi}{p} (-1)^p \frac{1-[\sqrt{1-4 z}]^{2p+1}}{2p+1} \right)=\\ &&\frac{1}{\sqrt{1-4z}} \cdot \left(\frac{1-\sqrt{1-4 z}}{2 z}\right)^{2j+1} \end{eqnarray}
Now, let us take arbitrary integer value of $j$. We use induction to prove the result. Clearly we have: \begin{eqnarray} rhs^{(j)}&=& \sum\limits_{i=0}^\infty \binom{2 i+j-1}{i} \cdot \underbrace{\frac{2i +j}{i+j}}_{2-\frac{j}{i+j}} \cdot z^i \\ &=& 2 \cdot\frac{1}{\sqrt{1-4 z}}\cdot \left( \frac{1-\sqrt{1-4 z}}{2 z}\right)^{j-1} - j \int\limits_0^1 \theta^{j-1} \frac{1}{\sqrt{1-4 z \theta}}\cdot \left( \frac{1-\sqrt{1-4 z \theta}}{2 z \theta}\right)^{j-1} d\theta\\ &=&2 \cdot\frac{1}{\sqrt{1-4 z}}\cdot \left( \frac{1-\sqrt{1-4 z}}{2 z}\right)^{j-1} +\frac{j}{(2 z)^j} \int\limits_1^{\sqrt{1-4z}} (1-u)^{j-1} du\\ &=&2 \cdot\frac{1}{\sqrt{1-4 z}}\cdot \left( \frac{1-\sqrt{1-4 z}}{2 z}\right)^{j-1} - \left(\frac{1-\sqrt{1-4 z}}{2 z} \right)^j\\ &=&\frac{1}{\sqrt{1-4z}} \cdot \left(\frac{1-\sqrt{1-4 z}}{2 z}\right)^{j} \end{eqnarray} as should be.
We consider the generating functions \begin{align*} A(z)=\frac{1}{\sqrt{1-4z}}\quad\text{and}\qquad B_2(z)=\frac{1-\sqrt{1-4z}}{2z} \end{align*} and derive for non-negative integer $r$ the coefficients of
\begin{align*} \left(B_2(z)\right)^r\qquad\text{ and }\qquad A(z)\left(B_2(z)\right)^r \end{align*}
We apply the following
Change of variable formula: Let $f(z)$ be a Laurent series and $g(w)$ be a power series, $g(w)=g_1w+g_2w^2+\cdots$, where $g_1\ne 0$. Then \begin{align*} \color{blue}{[z^{-1}]f(z)=[w^{-1}]f(g(w))g^\prime(w)}\tag{1} \end{align*}
See e.g. p.12 of this presentation by Ira Gessel.
Coefficients of $\left(B_2(z)\right)^r$: We use the transformation \begin{align*} z=\frac{w}{(1+w)^2}\qquad\qquad\frac{dz}{dw}=\frac{1-w}{(1+w)^3}\tag{2} \end{align*}
and we get \begin{align*} B_2(w)=\frac{1-\sqrt{1-\frac{4w}{(1+w)^2}}}{\frac{2w}{(1+w)^2}}=1+w \end{align*}
We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ a generating function $A(z)$ and obtain with (1) and (2) \begin{align*} \color{blue}{[z^n]\left(B_2(z)\right)^r}&=[z^{-1}]z^{-n-1}\left(\frac{1-\sqrt{1-4z}}{2z}\right)^r\\ &=[w^{-1}]\left(\frac{w}{(1+w)^2}\right)^{-n-1}(1+w)^r\frac{1-w}{(1+w)^3}\tag{3}\\ &=[w^n](1+w)^{2n+r-1}(1-w)\tag{4}\\ &=[w^n](1+w)^{2n+r-1}-[w^{n-1}](1+w)^{2n+r-1}\tag{5}\\ &=\binom{2n+r-1}{n}-\binom{2n+r-1}{n-1}\tag{6}\\ &=\binom{2n+r}{n}-2\binom{2n+r-1}{n-1}\tag{7}\\ &=\color{blue}{\binom{2n+r}{n}\frac{r}{2n+r}}\tag{8} \end{align*} and the claim follows. The coefficients $\binom{2n+1}{n}\frac{1}{2n+1}$ of $B_2(z)$ follow by setting $r=1$.
Comment:
In (3) we use the substitution (2).
In (4) we do some simplifications.
In (5) we factor out and apply the rule $[z^p]z^{q}A(z)=[z^{p-q}]A(z)$.
In (6) we select the coefficients accordingly.
In (7) we use the binomial identity $\binom{p+1}{q}=\binom{p}{q}+\binom{p}{q-1}$.
In (8) we do some final simplifications.
The same technique works for the other case.
Coefficients of $A(z)\left(B_2(z)\right)^r$:
We use the same transformation $z=\frac{w}{(1+w)^2}$ and we get \begin{align*} A(w)=\frac{1}{\sqrt{1-\frac{4w}{(1+w)^2}}}=\frac{1+w}{1-w} \end{align*}
We obtain with (1) and (2) \begin{align*} \color{blue}{[z^n]\left(A(z)\left(B_2(z)\right)^r\right)}&=[z^{-1}]z^{-n-1}\frac{1}{\sqrt{1-4z}}\left(\frac{1-\sqrt{1-4z}}{2z}\right)^r\\ &=[w^{-1}]\left(\frac{w}{(1+w)^2}\right)^{-n-1}\frac{1+w}{1-w}(1+w)^r\frac{1-w}{(1+w)^3}\\ &=[w^n](1+w)^{2n+r}\\ &=\color{blue}{\binom{2n+r}{n}} \end{align*} end the claim follows. The coefficients $\binom{2n}{n}$ of $A(z)$ follow by setting $r=0$.
