closed form for some binomial sum I am trying to derive a closed form for the generating function of
$a_n(x)=\sum_{k=0}^n \binom{n+k}{n}x^k,  x>0,  n\in\mathbb{N},$
i.e. for $G(z)=\sum_{n=0}^\infty a_n(x)z^n$. The only method I know is to establish a recurrence equation for the sequence $a_n(x)$. However I am failing to do so. Any advice or suggestion of different methods?
 A: Ok, I have to say, it is complicated but it works (see the final result before going through the whole calculus). You might want to have pages 52, 53 and 54 of this book : https://www.math.upenn.edu/~wilf/DownldGF.html in front of you. First set :
$$G(x,z):=\sum_{n=0}^{\infty}a_n(x)z^n $$
Furthermore :
$$F(x,z):=\sum_{n=0}^{\infty}\sum_{k=n+1}^{\infty}\binom{n+k}kx^kz^n  $$
We have :
$$G(x,z)+F(x,z)=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\binom{n+k}kx^kz^n $$
Now we reverse the order of summation to have :
$$G(x,z)+F(x,z)=\sum_{k=0}^{\infty}\left[\sum_{n=0}^{\infty}\binom{n+k}kz^n\right]x^k $$
Now use (2.5.7) (in the reference I gave) :
$$G(x,z)+F(x,z)=\sum_{k=0}^{\infty}\frac{x^k}{(1-z)^{k+1}}=\frac{1}{1-z}\frac{1}{1-\frac{x}{1-z}}=\frac{1}{1-(x+z)}$$
Now the hard part begins. I claim that I can find a closed form for $F(x,z)$. Remark that :
$$\binom{n+k}k=\binom{n+k}n\;,$$
hence:
$$F(x,z)=\sum_{n=0}^{\infty}\sum_{k=n+1}^{\infty}\binom{n+k}nx^kz^n=\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\binom{2n+k}nx^k(xz)^n$$
In the second equality I changed $k$ to $k:=k-n$. Now interchange the two sums :
$$F(x,z)=\sum_{k=1}^{\infty}\sum_{n=0}^{\infty}\binom{2n+k}nx^k(xz)^n $$
Use now (2.5.15) (in the reference I gave), setting $T:=\sqrt{1-4xz}$ :
$$F(x,z)=\sum_{k=1}^{\infty}x^k\frac{1}{T}\left(\frac{1-T}{2xz}\right)^k$$
$$F(x,z)=\frac{1}{T}\sum_{k=1}^{\infty}\left[\frac{1-T}{2z}\right]^k$$
$$F(x,z)=\frac{1}{T}\frac{\frac{1-T}{2z}}{1-\frac{1-T}{2z}}=\frac{1-T}{2zT-(T-T^2)}$$
Finally this is (kind of) a closed form for $F(x,z)$. We have that :
$$G(x,z)=\frac{1}{1-(x+z)}-\frac{1-\sqrt{1-4xz}}{2z\sqrt{1-4xz}-(\sqrt{1-4xz}-1+4xz)} $$
