Proof of identity about generalized binomial sequences. I was going through this old question about a wealthy gambler: 
Gambler with infinite bankroll reaching his target. The answer relies on the following identities from Concrete Mathematics by Graham, Knuth and Patashnik (equation numbers as they appear in the book).
$$B_2(z) = \sum\limits_{t=0}^\infty \frac{{2t+1\choose t}}{2t+1} z^t = \frac{1-\sqrt{1-4z}}{2z} \tag{5.68}$$
$$(B_2(z))^k = \left(\sum\limits_{t=0}^\infty {2t+1\choose t}\frac{1}{2t+1} z^t\right)^k = \sum\limits_{t=0}^\infty {2t+k \choose t} \frac{k}{2t+k}z^t \tag{5.70}$$
The expression on the far right of (5.70) is particularly interesting since  it is the stopping time of a wealthy gambler targeting $k$. 
It is also fascinating since $k$ seems to simply march into the infinite summation and replace $1$, somehow taking care of all the cross terms in the process. 
I read through the chapter to see if I could find a proof for these identities (both of which I verified numerically). 
Tracing my way back, I found the following (equivalent) definition of $B_u(z)$.
$$B_u(z) = \sum\limits_{t=0}^\infty \frac{ut \choose t}{(u-1)t+1} z^t \tag{5.58}$$
Then they simply state:
$$(B_u(z))^k = \sum\limits_{t=0}^\infty {ut+k \choose t} \frac{k}{ut+k} z^t \tag{5.60}$$
However, no proof is provided for these. So, I'm still scratching my head wondering how to prove (5.68) and (5.70).

My attempts:
For (5.70), we can say that in order for the gambler to reach $k$\$, he has to first reach $1$\$ and then repeat that feat $k$ times. This provides a rough sketch, but I'm still fascinated by the mechanical details (and (5.60) has no such interpretation in terms of gamblers). 
For (5.68), I tried some of the approaches in the answers to this question. 
First, Mathematica couldn't find a nice expression for the partial summation. So, @robojohn's approach probably won't work because if there were a function whose diff made up the terms of $B_2(z)$, the partial summation would have a nice expression in terms of that function.
Next, I tried @Marcus Scheuer's approach and got:
$$\frac{a_{t+1}}{a_t} = \frac{t+\frac 1 2}{t+2}(4z) = \frac{\frac{-1}{2}^\underline{t}}{-2^\underline{t}} (4z)$$
This doesn't work either since we don't get the $a+b=c+d$ condition required for the corollary he used and the $4z$ term interferes as well.
 A: Here is another approach I came across thanks to /u/whatkindofred on this reddit thread for proving (5.68). This approach starts from the LHS.
Let's suppose:
$$F(z) = \sum\limits_{t=0}^\infty a_t z_t =  \sum\limits_{t=0}^\infty \frac{2t \choose t}{t+1} z^t$$
It is easy to see that:
$$(t+1)a_t = (4t-2)a_{t-1}\tag{1.1}$$
Further suppose that:
$$G(z) = zF(z) = \sum\limits_{t=0}^\infty a_t z^{t+1}$$
So,
$$G'(z) = \sum\limits_{t=0}^\infty (t+1)a_t z^t$$
Using (1.1)
$$G'(z)= a_0 + \sum\limits_{t=1}^\infty(4t-2)a_{t-1}z^t$$
Since $a_0=1$,
$$G'(z) = 1+4 \sum\limits_{t=1}^\infty t a_{t-1} z^t - 2 \sum\limits_{t=1}^\infty a_{t-1}z_t$$
$$= 1+ 4 \sum_{t=1}^\infty (t+1)a_t z^{t+1} - 2 \sum\limits_{t=1}^\infty a_{t-1}z^t$$
$$G'(z)= 1+4zG'(z)-2G(z)\tag{1.2}$$
But since $G(z)=zF(z)$, 
$$G'(z)=F(z)+zF'(z)$$
Substituting into (1.2) we get:
$$F(z)+zF'(z)=1+2zF(z)+4z^2F'(z)$$
$$(4z^2-z)F'(z)+(2z-1)F(z)+1=0$$
$$F'(z) + g(z) F(z) = h(z) \tag{1.3}$$
Where,
$$g(z) = \frac{2z-1}{4z^2-z}$$
$$h(z)=\frac{1}{z-4z^2}$$
Multiplying both sides of (1.3) by $e^{\int\limits_{0}^x g(t)dt}$ we get,
$$e^{\int\limits_{0}^z g(t)dt} F'(z) + e^{\int\limits_{0}^x g(t)dt} g(z)F(z)=h(z)e^{\int\limits_{0}^z g(t)dt}$$
$$=> \frac{\partial}{\partial z}\left(F(z)e^{\int\limits_{0}^z g(t)dt}\right) = h(z) e^{\int\limits_{0}^z g(t)dt}$$
$$=> F(z)e^{\int\limits_{0}^z g(t)dt} = \int\limits_{y=0}^z h(y) e^{\int\limits_{0}^y g(t)dt}\tag{1.4}$$
Now, let's address the integrals.
$$\int g(z)dz = -\int \frac{2z-1}{z-4z^2}$$
$$ = \int \frac{-2}{1-4z}dz + \int \frac{dz}{z(1-4z)}$$
$$=\frac{\log(1-4z)}{2} + \int \frac{4z+(1-4z)}{z(1-4z)}dz$$
$$=\frac{\log(1-4z)}{2}+ 4 \int \frac{dz}{1-4z}+\int \frac{dz}{z}$$
$$=\frac{\log(1-4z)}{2}- \log(1-4z) +\log(z)$$
$$=> \int g(z) dz = \log\left(\frac{z}{\sqrt{1-4z}}\right)+b_1 $$
And so,
$$e^{\int g(z)dz} = c_1\frac{z}{\sqrt{1-4z}}\tag{1.5}$$
And this means,
$$\int h(z) e^{\int g(z)dz} = \int \frac{1}{z(1-4z)} c_2\frac{z}{\sqrt{1-4z}}dz$$
$$ = \int c_2(1-4z)^{-\frac 3 2}dz = \frac{c_2}{\sqrt{1-4z}}+c_3\tag{1.6}$$
Substituting (1.5) and (1.6) into (1.4) yields:
$$F(z)=\frac{d_1 + d_2 \sqrt{1-4z}}{z}$$
But we know that $F(0)=1$ and for the above equation to not blow up at $z=0$ we must have $d_1=-d_2=d$ giving us,
$$F(z) = d \left(\frac{1-\sqrt{1-4z}}{z}\right)$$
And using $\lim_{z \to 0}F(z)=1$ we get $d=\frac{1}{2}$ (use L' Hospitals rule) and the RHS of (5.68) follows.
A: Another easy way to see this is that if we substitute $z=p(1-p)$ in (5.68), the expression becomes the probability that the wealthy gambler will ever reach $k$\$ while (5.67) is the probability he will ever reach $1$\$ (if he keeps tossing a coin with probability $p$ of heads and wins $1$\$ on heads and loses $1$\$ on tails). To reach $k$\$, he has to increase his fortune by $1$\$ $k$ times. And the result follows.
