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

Question

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.

Answer 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 $$\frac{A}{B}=\frac{x}{(2x-2)(2x-1)}\cdot(4(x-1))=\frac{2x}{2x-1}$$

Answer 2

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)