Proof that ${2x\over 2x-1}={A(x)\over A(x-1)}$ for every integer $x\ge2$, where $A(x)=\sum\limits_{n=0}^\infty\prod\limits_{k=0}^n\frac{x-k}{x+k}$ 
Let $x\ge2$ denote an integer. Consider:
  $$A=1+{x-1\over x+1}+{(x-1)(x-2)\over (x+1)(x+2)}+{(x-1)(x-2)(x-3)\over (x+1)(x+2)(x+3)}+\cdots\tag1$$
  and 
  $$B=1+{x-2\over x}+{(x-2)(x-3)\over x(x+1)}+{(x-2)(x-3)(x-4)\over x(x+1)(x+2)}+\cdots\tag2$$
  How does one show that $${2x\over 2x-1}={A\over B}\ ?$$

An attempt: consider $(x)^n=x(x+1)\cdots(x+n-1)$ and $(x)_n=x(x-1)\cdots(x-(n-1))$. One can re-write $(1)$ as
$$A=x+{(x)_2\over (x)^2}+{(x)_3\over (x)^3}+{(x)_4\over (x)^4}+\cdots\tag3$$
and $(2)$ as
$$x+x(x-1)B=x+(x)_2+{(x)_3\over x}+{(x)_4\over (x)^2}+{(x)_5\over (x)^3}+\cdots\tag4$$
but I am not sure how to continue.

Frist
$$
\prod^{n}_{k=0}\frac{x-k}{x+k}=\prod^{n}_{k=1}\frac{x-k}{x+k}=(-1)^n\frac{\prod^{n}_{k=1}(-x+1+k)}{\prod^{n}_{k=1}(x+1+k)}=(-1)^n\frac{(-x+1)_n}{(x+1)_n},
$$
where $(a)_n:=a(a+1)(a+2)\ldots(a+n-1)$. Hence
$$
A(x)=\sum^{\infty}_{n=0}(-1)^n\frac{(1-x)_n}{(x+1)_n}=\sum^{\infty}_{n=0}\frac{(1-x)_n(1)_n}{(x+1)_n}\frac{(-1)^n}{n!},
$$
since $(1)_n=n!$, for $n=0,1,2,\ldots$. Hence
$$
A(x)={}_2F_{1}\left(1-x,1;x+1;-1\right) \tag 1
$$
But from [Leb] Chapter 9, pg. 240 relation (9.1.6) it is known that
$$
{}_2F_1\left(a,b;c;z\right)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int^{1}_{0}t^{b-1}(1-t)^{c-b-1}(1-tz)^{-a}dt,
$$
where $Re(c)>Re(b)$, $|\textrm{arg}(1-z)|<\pi$. 
Hence whenever $x>0$:
$$
A(x)=\frac{\Gamma(x+1)}{\Gamma(x)}\int^{1}_{0}(1-t)^{x-1}(1+t)^{x-1}dt=x\int^{1}_{0}(1-t^2)^{x-1}dt=
$$
$$
=\frac{x}{2}\int^{1}_{0}(1-w)^{x-1}w^{-1/2}dw
=\frac{x}{2}B\left(x,\frac{1}{2}\right)=\frac{x}{2}\frac{\Gamma(x)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(x+\frac{1}{2}\right)}=\frac{1}{2}\frac{\Gamma(x+1)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(x+\frac{1}{2}\right)}\tag 2
$$
Where we have make the change of variable $x\rightarrow\sqrt{w}$ and use the identity (see [Leb] Chapter 1, pg.13-14)
$$
\int^{1}_{0}t^{a_1-1}(1-t)^{b_1-1}dt=\frac{\Gamma(a_1)\Gamma(b_1)}{\Gamma(a_1+b_1)},
$$
where $Re(a_1)>0$, $Re(b_1)>0$. Hence
$$
\frac{A(x)}{A(x-1)}=\frac{\Gamma(x+1)\Gamma\left(x-\frac{1}{2}\right)}{\Gamma(x)\Gamma\left(x+\frac{1}{2}\right)}=\frac{x\Gamma\left(x-\frac{1}{2}\right)}{\left(x-\frac{1}{2}\right)\Gamma\left(x-\frac{1}{2}\right)}=\frac{2x}{2x-1}\tag 3
$$
where we have used the identity $\Gamma(x+1)=x\Gamma(x)$, $x>0$.
Hence we have proved that
$$
\frac{A(x)}{A(x-1)}=\frac{2x}{2x-1},
$$
$x$ real greater than 1.
References
[Leb] N.N. Lebedev "Special Functions and their Applications". Dover Publications, Inc. New York (1972)
