Prove the following binomial identity $$\sum_{i=1}^{n-1}\frac {\displaystyle\binom {2i-1}i}{2i-1}\cdot \frac {\displaystyle\binom {2n-2i-1}{n-i}}{2n-2i-1}=\frac {\displaystyle\binom {2n-1}n}{2n-1}
$$
I tried the following method:
Let $$A(x,y)=\sum^{n-1}_{i=1}\frac{(x+\frac{1}{x})^{2i-1}}{2i-1}\cdot \frac{(y+\frac{1}{y})^{2n-2i-1}}{2n-2i-1}$$
One can analyze the coefficient for the term $xy$, which is exactly LHS. 
One can also do the operation $\frac{\partial}{\partial x}\frac{\partial}{\partial y}$ first and then analyze the constant term, which should be the RHS. However, the problem is that the constant term is not so easy to find and I want to ask whether someone can continue with my proof or, otherwise, provide other methods to approach this problem. Thank you!
 A: First, we note that if we define
$$
a_i = \frac{\binom{2i-1}{i}}{2i-1}
$$
for $i\ge1$, $a_i=0$ otherwise, the equation reads $a_n=\sum_{i=1}^{n-1}a_ia_{n-i}$ for $n\ge2$. Note that it does not hold for $n=1$.
Let's first simplify $a_i$ a bit by using that $(2k)!/k!=2^k\cdot1\cdot3\cdots(2k-1)$:
$$
a_i = \frac{\binom{2i-1}{i}}{2i-1}
= \frac{(2i-2)!}{i!(i-1)!}
= \frac{2^{i-1}\cdot1\cdot3\cdots(2i-3)}{i!}
= \frac{4^{i-1}}{i!}\cdot\frac{1}{2}\cdot\frac{3}{2}\cdots\frac{2i-3}{2}
$$
with $a_1=1$.
There are a few ways to proceed from here. One is to notice that this can also be written as $a_i=2\cdot(-4)^{i-1}\cdot\binom{1/2}{i}$ for $i\ge1$, using the definition $\binom{s}{i}=s(s-1)\cdots(s-i+1)/i!$ where $i$ is a non-negative integer while $s$ may be any real or complex number. Note, however, that while $a_0=0$, the expression $2\cdot(-4)^{i-1}\cdot\binom{1/2}{i}$ for $i=0$ becomes $-1/2$. Thus,
$$
\sum_{i=1}^{n-1} a_ia_{n-i}
= \sum_{i=0}^n 4\cdot(-4)^{n-2}\binom{1/2}{i}\binom{1/2}{n-i} + a_n
= (-1)^n 4^{n-1}\binom{1}{n} + a_n
= a_n
$$
for $n\ge2$.
Another approach is to note that
$$
a(x) = \sum_{i=1}^\infty a_ix^i = \frac{1-\sqrt{1-4x}}{2}
$$
and
$$
\sum_{n=2}^\infty\sum_{i=1}^{n-1} a_ia_{n-i}x^n
= \sum_{i=1}^\infty a_ix^i \cdot \sum_{j=1}^\infty a_jx^j
= a(x)^2
= \frac{1-2\sqrt{1-4x}+1-4x}{4}
= a(x)-x
$$
which gives the coefficients $a_n$ for $n\ge2$. 
