<s>[INST] $(\frac1a+\frac12\frac{x}{a+2}+\frac{1\cdot3}{2\cdot4}\frac{x^2}{a+4}+...)(1+\frac12x+\frac{1\cdot3}{2\cdot4}x^2+...)=\frac1a(1+\frac{a+1}{a+2}x+...)$ For $a>0$, prove $$\left(\frac{1}{a}+\frac{1}{2}\cdot\frac{x}{a+2}+\frac{1\cdot 3}{2\cdot 4}\cdot \frac{x^2}{a+4}+\cdots\right) \cdot 
\left( 1+\frac{1}{2}\cdot x+\frac{1\cdot 3}{2\cdot 4}\cdot  x^2 +\cdots \right)
= \frac{1}{a} \left[
1+\frac{a+1}{a+2}x+\frac{(a+1)(a+3)}{(a+2)(a+4)} x^2 +\cdots
\right]$$
I'm a bit lost on how to prove it.
First I tried to prove it by induction but the expansion of $x^n$'s coefficient on the left side is complex.
Then I tried to simplify it. Call it $S_1 S_2 = S_3$,
apparently $S_2 = \sum_{n\ge 0} 4^{-n}{2n \choose n} x^n  = \frac{1}{\sqrt{1-x}}$, but I got problem in $S_1$.
What I have done is:
Denote $c_n = 4^{-n} {2n \choose n}$, then
$$\left( \sum_{n\ge 1} \frac{c_n}{n} x^n \right)' = \sum_{n\ge 1} c_n x^{n-1} = \frac1x \left( \sum_{n\ge 0} c_n x^n - 1 \right) = \frac1x\left(\frac{1}{\sqrt{1-x}}-1\right) $$
Let $f(x) = -2 \ln (1+\sqrt{1-x})$, then $f' = \frac1x\left(\frac{1}{\sqrt{1-x}}-1\right)$.
Integral from $0$ to $x$ leads to
$$\sum_{n\ge 1} \frac{c_n}{n} x^n  = \int_0^x \frac1x\left(\frac{1}{\sqrt{1-x}}-1\right) dx = f(x) |_0^x =\ln 4 - 2 \ln (1+\sqrt{1-x})$$
To work out $S_1=\sum_{n\ge 1} \frac{c_n}{2n+a} x^n = \frac12 \sum_{n\ge 1} \frac{c_n}{n+b} x^n$, where $b = a/2$, define $g(x) = 2S_1 = \sum_{n\ge 1} \frac{c_n}{n+b} x^n $. Then we have
$$ \frac{(x^b g)'}{x^b} = g' + \frac{b}{x}g = \frac1x\left(\frac{1}{\sqrt{1-x}}-1\right) $$
So
$$g = \frac{1}{x^b} \left( \int_0^x \frac{x^{b-1}}{\sqrt{1-x}} dx - \frac{x^b}{b} + c  \right) $$
, where $c$ is a constant.
But then I'm stuck at the ​$ \int_0^x \frac{x^{b-1}}{\sqrt{1-x}} dx$
integral.
 [/INST] After forming the Cauchy product for the LHS, we see that it suffices to prove that $$\sum_{k=0}^n\frac{c_k c_{n-k}}{a+2k}=\frac1a\prod_{k=1}^n\frac{a+2k-1}{a+2k}\tag{*}\label{coeffs}$$ where $c_n=\displaystyle\prod_{k=1}^n\frac{2k-1}{2k}$ as in the OP.
But \eqref{coeffs} is just the partial fraction decomposition of its RHS: $$f(a):=\frac1a\prod_{k=1}^n\frac{a+2k-1}{a+2k}=\sum_{k=0}^n\frac{b_k}{a+2k},$$ where $b_k$ are constants (that don't depend on $a$, and) determined by $$b_j=\lim_{a\to-2j}(a+2j)f(a)=\frac{\prod_{1\leqslant k\leqslant n}(2k-2j-1)}{\prod_{0\leqslant k\leqslant n,\ k\neq j}(2k-2j)}=c_j c_{n-j}$$ after writing the numerator as $\prod_{k=1}^j\cdot\prod_{k=j+1}^n$ and the denominator as $\prod_{k=0}^{j-1}\cdot\prod_{k=j+1}^n$.
</s>