Proof for the identity $ \sum _{k=1}^j \frac{1}{2 k-1} \binom{2 k}{k} \binom{2 j-2 k}{j-k} =\binom{2 j}{j} $ I know that the following identity is true (e.g., using Mathematica)
$$
\sum _{k=1}^j \frac{1}{2 k-1} \binom{2 k}{k} \binom{2 j-2 k}{j-k} =\binom{2 j}{j}
$$
but I would like a proof of it, by any mean. I have tried by induction or with the following polynomial
$$
\sum _{k=1}^j \binom{2 k}{k} \binom{2 j-2 k}{j-k} x^{2k-2}
$$
with the intent of integrating it, but I had no luck in either case.
Thanks a lot for your help.
 A: Given any infinite sequence of numbers
$\,\{t_0,t_1,\dots\},\,$ define the sequences
$$ a_n := t_n, \quad
   b_n := \frac{t_n}{2n-1}, \quad
   c_n := \sum_{k=1}^n a_{n-k}b_k. \tag{1} $$
Define their corresponding power series generating functions
$$ A := \sum_{n=0}^\infty a_n x^n, \quad
   B := \sum_{n=1}^\infty b_n x^n, \quad
   C := \sum_{n=1}^\infty c_n x^n. \tag{2} $$
Use the definitions in equation $(1)$ to get $\,A\,B = C.\,$
Define
$$ t_n := {2n\choose n}\qquad \text{ and }
\qquad y := \sqrt{1-4x}. \tag{3}$$
Use binomial coefficients results (
OEIS A000984 and
OEIS A002420) to get
$$ A = \frac1y \qquad \text{ and } \qquad B = 1-y. \tag{4} $$
Multiply to get
$$ A\,B = \frac1y(1-y) = \frac1y - 1 = A-1. \tag{5} $$
Use $\,A\,B = C,\,$ to get
$\,c_n = a_n\,$ for all $\,n>0\,$
which proves the identity.
A: To prove such identities, I often prefer combinatorial approaches like "Double Counting" rather than "Snake Oil" or "WZ Pairs" methods. We will count the below quantity in two methods:
The number of permutations of $j$ digits $0$ and $j$ digits $1$.

*

*First method: Consider $2j$ places in a row. To create a such permutation, we can choose $j$ places and put $0$'s in these places. The rest places are owned by the $1$'s. So the number of ways of the first method is $\binom{2j}{j}$.


*Second method: For each $1 \leq k\leq j$, we count
the permutations of $j$ digits $0$ and $j$ digits $1$ such that "the number of $0$'s = the number of $1$'s" occurs for the first time at index $2k$.
It's easy to verify that when $k$ varies from $1$ to $j$, the above permutations contain all of the desired permutations, and they don't intersect each other. So let's count the number of such permutations for a fixed $k$. Suppose that there exist $a_k$ ways for the first $2k$ places. For the rest $2j - 2k$ places, we can choose $j - k$ of them and put $0$'s into them. Then the place of $1$'s will be determined uniquely. So the number of such permutations for a fixed $k$ is $a_k\binom{2j-2k}{j-k}$. Therefore, to complete the proof we only must show that $a_k = \frac{1}{2k-1}\binom{2k}{k}$.
Suppose we are at the point $(0, 0)$ in the coordinate plane and for each digit $0$, we go a unit to the right and for each $1$, we go a unit to the up. So $a_k$ equals the number of ways we can reach $(k, k)$ such that we always walk strictly below the line $y = x$ or always strictly above this line. Oh! The famous Catalan numbers appeared! One can easily see that the situation is equivalent to going from the point $(0, 1)$ to $(k, k-1)$ (or $(1, 0)$ to $(k-1, k)$) such that we always walk below (not necessary "strictly") the line $y = x - 1$ (or always above the line $y = x+1$). So $a_k = 2C_{k-1} = 2 \times \frac{1}{k}\binom{2k-2}{k-1} = \frac{1}{2k-1}\binom{2k}{k}$.

Source of the picture
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\begin{align}
&\bbox[5px,#ddf]{\left.\quad\sum_{k = 1}^{j}{1 \over 2k - 1}{2k \choose k}{2j - 2k \choose j - k}
\right\vert_{\, j\ \in\ \mathbb{N}_{\,\geq\ 1}}\quad }
\\[5mm] = & \
\sum_{k = 1}^{j}\pars{\int_{0}^{1}t^{2k - 2}\,\,\dd t}
\bracks{{-1/2 \choose k}\pars{-4}^{k}}
\bracks{{-1/2 \choose j - k}\pars{-4}^{j - k}}
\\[5mm] = & \
\pars{-4}^{j}\int_{0}^{1}\sum_{k = 1}^{\infty}
{-1/2 \choose k}{-1/2 \choose j - k}t^{2k - 2}\,\,\dd t
\\[5mm] = & \
\pars{-4}^{j}\int_{0}^{1}\sum_{k = 1}^{\infty}
{-1/2 \choose k}
\braces{\bracks{z^{j - k}}\pars{1 + z}^{-1/2}}
t^{2k - 2}\,\,\dd t
\\[5mm] = & \
\pars{-4}^{j}\bracks{z^{j}}\pars{1 + z}^{-1/2}\int_{0}^{1}\sum_{k = 1}^{\infty}
{-1/2 \choose k}\pars{zt^{2}}^{k}
\,\,{\dd t \over t^{2}}
\\[5mm] = & \
\pars{-4}^{j}\bracks{z^{j}}\pars{1 + z}^{-1/2}\int_{0}^{1}
{\pars{1 + zt^{2}}^{-1/2} - 1 \over t^{2}}\,\dd t
\\[5mm] = & \
\pars{-4}^{j}\bracks{z^{j}}\pars{1 + z}^{-1/2}
\bracks{1 - \pars{1 + z}^{1/2}}
\\[5mm] = & \
\pars{-4}^{j}\braces{\bracks{z^{j}}\pars{1 + z}^{-1/2}
- \bracks{z^{j}}1}
\\[5mm] = & \
\pars{-4}^{j}{-1/2 \choose j} =
\pars{-4}^{j}{{2j \choose j} \over \pars{-4}^{j}}=
\bbx{\large2j \choose j}
\end{align}
