Proof that ${2x\over 2x-1}={A(x)\over A(x-1)}$ for every integer $x\ge2$, where $A(x)=\sum\limits_{n=0}^\infty\prod\limits_{k=0}^n\frac{x-k}{x+k}$

Solution 1:

Lets get every term in $A$ and in $B$ to the same denominator,using your notation.The common denominator is finite since after $(2x-1)$ in $A$ the numerator perishes,and in $B$ after $(2x-3)$ the numerator perishes. $$A=\frac{(x+1)^{x-1}+(x-1)_1(x+2)^{x-2}+(x-1)_2(x+3)^{x-3}+\cdots+(x-1)_{x-1}}{(x+1)^{x-1}}\\B=\frac{(x)^{x-2}+(x-2)_1(x+1)^{x-3}+(x-2)_2(x+2)^{x-4}+\cdots+(x-2)_{x-2}}{x^{x-2}}$$From this we have that$$\frac{A}{B}=\frac{x^{x-2}}{(x+1)^{x-1}}\cdot\frac{(x+1)^{x-1}+(x-1)_1(x+2)^{x-2}+(x-1)_2(x+3)^{x-3}+\cdots+(x-1)_{x-1}}{(x)^{x-2}+(x-2)_1(x+1)^{x-3}+(x-2)_2(x+2)^{x-4}+\cdots+(x-2)_{x-2}}\\\frac{A}{B}=\frac{x}{(2x-2)(2x-1)}\cdot\frac{(x+1)^{x-1}+(x-1)_1(x+2)^{x-2}+(x-1)_2(x+3)^{x-3}+\cdots+(x-1)_{x-1}}{(x)^{x-2}+(x-2)_1(x+1)^{x-3}+(x-2)_2(x+2)^{x-4}+\cdots+(x-2)_{x-2}}$$ Now lets look at $$C=\frac{(x+1)^{x-1}+(x-1)_1(x+2)^{x-2}+(x-1)_2(x+3)^{x-3}+\cdots+(x-1)_{x-1}}{(x)^{x-2}+(x-2)_1(x+1)^{x-3}+(x-2)_2(x+2)^{x-4}+\cdots+(x-2)_{x-2}}$$ We have that $$(x)^n=\frac{(x+n-1)!}{(x-1)!},(x)_n=\frac{x!}{(x-n)!}$$ Applying the two formulas we get $$C=\frac{\frac{(2x-1)!}{x!}+\frac{(2x-1)!(x-1)!}{(x+1)!(x-2)!}+\frac{(2x-1)!(x-1)!}{(x+2)!(x-3)!}+\cdots+\frac{(2x-1)!(x-1)!}{(2x-1)!}}{\frac{(2x-3)!}{(x-1)!}+\frac{(2x-3)!(x-2)!}{x!(x-3)!}+\frac{(2x-3)!(x-2)!}{(x+1)!(x-4)!}+\cdots+\frac{(2x-3)!(x-2)!}{(2x-3)!}}=\frac{(x-1)!}{(x-2)!}\cdot\frac{{2x-1\choose x}+{2x-1\choose x+1}+\cdots+{2x-1\choose 2x-2}+{2x-1\choose 2x-1}}{{2x-3\choose x-1}+{2x-3\choose x}+\cdots+{2x-3\choose 2x-4}+{2x-3\choose 2x-3}}$$ Since$${2x-1\choose x}+{2x-1\choose x+1}+\cdots+{2x-1\choose 2x-2}+{2x-1\choose 2x-1}={2x-1\choose 0}+{2x-1\choose 1}+\cdots+{2x-1\choose x-2}+{2x-1\choose x-1}$$ We have that $${2x-1\choose x}+{2x-1\choose x+1}+\cdots+{2x-1\choose 2x-2}+{2x-1\choose 2x-1}=2^{2x-2}$$ Similarly we can deduce that $${{2x-3\choose x-1}+{2x-3\choose x}+\cdots+{2x-3\choose 2x-4}+{2x-3\choose 2x-3}}=2^{2x-4}$$ From this we have that $C=4(x-1)$ so $$\frac{A}{B}=\frac{x}{(2x-2)(2x-1)}\cdot(4(x-1))=\frac{2x}{2x-1}$$