How to prove combinatorial identities related to infinite series? This is an identity for the sum of combinatorial numbers. What techniques are used to prove it?

\begin{align*}
\dfrac{1}{4^n}\sum _{i=0}^n \frac{1}{2 i+k}\binom{2 i}{i} \binom{2 n-2 i}{n-i}=\dfrac{(k-2)\text{!!}}{(k-1)\text{!!}} \cdot\frac{(2n+k-1)\text{!!}}{(2n+k)\text{!!}}
\end{align*}

I met this identity when expanding the series, but how to prove it?
\begin{align*}
S&=\dfrac{1}{k}\sum _{n=0}^{\infty}\frac{\prod _{j=1}^n (2 j+k-1)}{\prod _{j=1}^{n} (2j+k)}x^{2n+k}\\
&=\dfrac{(k-2)\text{!!}}{(k-1)\text{!!}}\sum _{n=0}^{\infty} \frac{(2n+k-1)\text{!!}}{(2n+k)\text{!!}}x^{2n+k}\\
&=\sum _{n=0}^{\infty} \frac{x^{2 n+k}}{4^n}\sum _{i=0}^n \frac{1}{2 i+k}\binom{2 i}{i} \binom{2 n-2 i}{n-i}\\
&=\left\{\sum _{m=0}^{\infty}\frac{(2m-1)\text{!!}}{\left[(2m)\text{!!}\right](2m+k)}
 x^{2m+k}\right\}\left[\sum _{m=0}^{\infty}\frac{(2m-1)\text{!!}}{(2m)\text{!!}}
 x^{2m}\right]\\
&=\left(\int_0^x \frac{t^{k-1}}{\sqrt{1-t^2}} \,\mathrm{d}t\right)\left(\dfrac{1}{\sqrt{1-x^2}}\right)\\
\end{align*}
 A: Probably not the answer you're looking for (which is why I planned to leave it as a comment, but it got too long) but your sum is hypergeometric in $i$, which means there's a computer algorithm that can compute a closed form for you. In fact, there are computer algorithms which will (almost always) give you a proof involving only basic algebra (though the resulting proof looks like magic, and would be almost impossible for a human to come up with!). Hopefully someone else comes along to give an analytic answer, but for now let's see how we can have a computer solve this problem for us:
First, let's talk about the sum. We can ask sage to compute
$$
\frac{1}{4^n} \sum_{j=0}^n \frac{1}{2j + k} \binom{2j}{j} \binom{2n-2j}{n-j}
$$
1/4^n * sum(1/(2*j + k) * binom(2*j, j) * binom(2*n - 2*j, n-j), j, 0, n)  

(where I'm using $j$ as the variable because sage thinks $i$ is the complex unit) and it happily outputs
factorial(1/2*k)*factorial(1/2*k + n - 1/2)/(k*factorial(1/2*k + n)*factorial(1/2*k - 1/2))

or, more legibly,
$$
\frac{\left(\frac{1}{2} \, k\right)! \left(\frac{1}{2} \, k + n - \frac{1}{2}\right)!}{k \left(\frac{1}{2} \, k + n\right)! \left(\frac{1}{2} \, k - \frac{1}{2}\right)!}
$$
which we can quickly massage into your expression by using the well known identity
$\left ( \frac{k}{2} \right )! = \frac{k!!}{2^k}$.
Now, the algorithm that sage uses has been proven correct (for instance, in the book A=B), so barring implementation bugs, this sum is correct, and you can carry on with your research.

However, I mentioned at the start of this answer that there are computer algorithms which can give you an easy proof of the claim, but that it will look like magic. The relevant search term is "WZ Certificate". There's also a nice discussion here and in chapter 7 of A=B.
For simplicity, let's look at the $k=2$ case. Then we're trying to prove
$$
\frac{1}{4^n} \sum_{j=0}^n \frac{1}{2j+2} \binom{2j}{j} \binom{2n-2j}{n-j} 
= \frac{(2n+1)!!}{(2n+2)!!}
$$
Equivalently, replacing the double factorials by single factorials and moving everything to the left hand side, we're trying to prove
$$
\sum_{j=0}^n F(n,j) 
= \sum_{j=0}^n \frac{(n+1) \binom{2j}{j} \binom{2n-2j}{n-j}}{(j+1) (2n+1) \binom{2n}{n}} 
= 1 \quad \quad (\star)
$$
now for the black magic: we can ask sage for the WZ certificate
F(n,j) = (n + 1)*binomial(2*j, j)*binomial(-2*j + 2*n, -j + n)/((j + 1)*(2*n + 1)*binomial(2*n, n))
F.WZ_certificate(n,j)

and sage tells us the certificate is
$$
R(n,j) = -\frac{{\left(2 \, j - 2 \, n - 1\right)} {\left(j + 1\right)} j}{{\left(j - n - 1\right)} {\left(2 \, n + 3\right)} {\left(n + 1\right)}}
$$
why care? Because even though we might not have been able to come up with $R$ by ourselves, once we have it it makes it almost trivial to prove $(\star)$! But since $\star$ is just a manipulation of our sum of interest, we'll be done. So let's see how to do it:

Let $G(n,j) = R(n,j) \cdot F(n,j)$. Then (tedious but) basic algebra shows that
$$G(n,j+1) - G(n,j) = F(n+1,j) - F(n,j)$$
if we sum this over all $j$, we see that
$$\sum_j G(n,j+1) - \sum_j G(n,j) = \sum_j F(n+1,j) - \sum_j F(n,j)$$
but, of course, the left hand side is $0$ since we're summing over all $j \in \mathbb{Z}$, and so the two $G$-sums are just reindexed versions of each other!
Then the right hand side is $0$ too, so $\sum_j F(n+1,j) = \sum_j F(n,j)$ and our sum is independent of $n$. But it's easy to check that $\sum_j F(0,j) = 1$, so we see that $\sum F(n,j) = 1$. As desired.

Unfortunately, sage is refusing to compute $R$ when we leave $k$ as a parameter, but if you really want a proof there's nothing stopping you from running the WZ algorithm by hand, keeping track of $k$ as you go. Alternatively, if you have access to mathematica or maple you can probably convince them to get the certificate for you.

I hope this helps ^_^
