TITLE: Prove $(1-x)^{2k+1} \sum\limits_{n\ge 0}\binom{n+k-1}{k}\binom{n+k}{k} x^n = {\sum\limits_{j\ge 0} \binom{k-1}{j-1}\binom{k+1}{j} x^j} $
QUESTION [7 upvotes]: I stumbled upon the identity
$$(1-x)^{2k+1} \sum\limits_{n\ge 0}\binom{n+k-1}{k}\binom{n+k}{k} x^n = {\sum\limits_{j\ge 0} \binom{k-1}{j-1}\binom{k+1}{j} x^j}. $$
The right-hand side is a polynomial. This is easy to verify for small $k$, but I don't see a simple proof for all $k$.
 Can anyone help?
Edit:
It should perhaps be noted that the right-hand side is essentially a Narayana polynomial  ${N_k}(x) = \sum\limits_{j = 1}^k \binom{k}{j-1}\binom{k}{j}/k x^j. $
More precisely ${\sum\limits_{j\ge 0} \binom{k-1}{j-1}\binom{k+1}{j} x^j}=(k+1) N_k(x).$
Edit after the answer by Kelenner:
After studying both ideas of Kelenner in more detail I have seen that both lead to a proof of the proposed identity if one is willing to accept computer proofs. As I have stated in a comment the first idea leads to the identity $$\sum\limits_{m = 0}^k \binom{k-1}{m}\binom{2k-m}{k}x^{k-m}(1-x)^m={\sum\limits_{j= 0}^k \binom{k-1}{j-1}\binom{k+1}{j} x^j}.$$ My first reaction was that this leads to ugly binomial sums and therefore at first I did not study it further. 
I prefer in general proofs by hand but then I was curious to see if it could be done by computer. And it worked. Comparing coefficients of $x^j$ this identity reduces to
$$\sum\limits_{m = 0}^k (-1)^{j-m}\binom{k-1}{k-m}\binom{k+m}{k}\binom{k-m}{j-m}=\binom{k-1}{j-1}\binom{k+1}{j}.$$
After dividing by the RHS we get a sum which should be the constant $1$.
To this sum I applied the Mathematica Fast Zeilberger Package by Peter Paule, Markus Schorn and Axel Riese and got the desired result.
In the same way the partial fraction identity for the Narayana polynomials can be proved.

REPLY [2 votes]: Suppose  we start  by  evalutating the  two  sums in  turn, where  the
parameter $k\ge 1$.
For the first we will be using the following integral representation:
$${n+k\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+k}}{z^{k+1}} \; dz.$$
We seek $$\sum_{n\ge 1} {n-1+k\choose k} {n+k\choose k} x^n.$$
Using the integral we find
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\sum_{n\ge 1} {n-1+k\choose k} x^n
\frac{(1+z)^{n+k}}{z^{k+1}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^k}{z^{k+1}}
\sum_{n\ge 1} {n-1+k\choose k} (1+z)^n x^n \; dz
\\= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{x(1+z)^{k+1}}{z^{k+1}}
\sum_{n\ge 1} {n-1+k\choose k} (1+z)^{n-1} x^{n-1} \; dz
\\= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{x(1+z)^{k+1}}{z^{k+1}}
\frac{1}{(1-x(1+z))^{k+1}} \; dz
\\= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{x(1+z)^{k+1}}{z^{k+1}}
\frac{1}{(1-x-xz))^{k+1}} \; dz
\\= \frac{1}{(1-x)^{k+1}} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{x(1+z)^{k+1}}{z^{k+1}}
\frac{1}{(1-xz/(1-x)))^{k+1}} \; dz
\\ = \frac{x}{(1-x)^{k+1}}
\sum_{q=0}^k {k+1\choose k-q} 
{q+k\choose k} \left( \frac{x}{1-x} \right)^q
\\ = \frac{x}{(1-x)^{k+1}}
\sum_{q=0}^k {k+1\choose q+1} 
{q+k\choose k} \left( \frac{x}{1-x} \right)^q.$$
Applying the integral representation from the beginning a second time
we obtain for this sum
$$\frac{x}{(1-x)^{k+1}} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{q=0}^k {k+1\choose q+1} 
\frac{(1+z)^{q+k}}{z^{k+1}} 
\left( \frac{x}{1-x} \right)^q\; dz
\\ = \frac{x}{(1-x)^{k+1}} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^k}{z^{k+1}}
\sum_{q=0}^k {k+1\choose q+1} 
(1+z)^q
\left( \frac{x}{1-x} \right)^q\; dz
\\ = \frac{1}{(1-x)^k} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{k-1}}{z^{k+1}}
\sum_{q=0}^k {k+1\choose q+1} 
(1+z)^{q+1}
\left( \frac{x}{1-x} \right)^{q+1}\; dz
\\ = \frac{1}{(1-x)^k} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{k-1}}{z^{k+1}}
\left(-1 + \left(1+(1+z)\frac{x}{1-x}\right)^{k+1}\right) \; dz.$$
We have $k+1-(k-1) = 2$, so the first component inside the parentheses
drops out, leaving
$$ \frac{1}{(1-x)^k} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{k-1}}{z^{k+1}}
\left(1+(1+z)\frac{x}{1-x}\right)^{k+1} \; dz
\\ = \frac{1}{(1-x)^{2k+1}} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{k-1}}{z^{k+1}}
\left(1-x+x(1+z)\right)^{k+1} \; dz
\\ = \frac{1}{(1-x)^{2k+1}} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{k-1}}{z^{k+1}}
\left(1+xz\right)^{k+1} \; dz.$$
We need one more simplification on this and put $z=1/w$, getting
$$\frac{1}{(1-x)^{2k+1}} \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+1/w)^{k-1}}{(1/w)^{k+1}}
\left(1+x/w\right)^{k+1} \; \frac{1}{w^2} dw
\\ = \frac{1}{(1-x)^{2k+1}} \frac{1}{2\pi i}
\int_{|w|=\epsilon} w^2 (w+1)^{k-1}
\left(\frac{w+x}{w} \right)^{k+1} \; \frac{1}{w^2} dw
\\ = \frac{1}{(1-x)^{2k+1}} \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(w+1)^{k-1}}{w^{k+1}} (w+x)^{k+1}\; dw.$$
The reson this works is because we are essentially evaluating the residue at infinity and the residues sum to zero.
This concludes the evaluation of the first sum. 
For the second we will be using the following integral representation:
$${k-1\choose j-1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{k-1}}{z^j} \; dz.$$
We seek $$\sum_{j\ge 1} {k+1\choose j} {k-1\choose j-1} x^j.$$
Using the integral we find
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\sum_{j\ge 1} {k+1\choose j} x^j
\frac{(1+z)^{k-1}}{z^j} \; dz
\\= \frac{1}{2\pi i}
\int_{|z|=\epsilon} (1+z)^{k-1}
\sum_{j\ge 1} {k+1\choose j} \frac{x^j}{z^j} \; dz
\\= \frac{1}{2\pi i}
\int_{|z|=\epsilon} (1+z)^{k-1}
\left(-1 + (1+x/z)^{k+1} \right) \; dz.$$
The entire component drops out, leaving
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} (1+z)^{k-1}
(1+x/z)^{k+1} \; dz
\\= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{k-1}}{z^{k+1}}
(z+x)^{k+1} \; dz.$$
This however is  precisely the integral that we had  for the first sum
without the factor in front, done.
The only infinite sum appearing here is the first one with convergence
when $|(1+z)x|<1.$  Therefore choosing  $|x|\lt 1/Q$ and  $|z|\lt 1/Q$
with  $Q\ge  2$  we  have  $|(Q+1)/Q/Q|=  |1/Q^2  +  1/Q|<1$  and  get
convergence of the first LHS integral in a neighborhood of zero.
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.