$C_{k}={1\over 1+k}{2k\choose k}$; it is the k-th Catalan numbers
$x\ge2$
$$-{C_{x-2}\over {2x-1\choose x}}=1-{x-1\over x+1}(2^2)+{(x-1)(x-2)\over (x+1)(x+2)}(3^2)-{(x-1)(x-2)(x-3)\over (x+1)(x+2)(x+3)}(4^2)+\cdots$$
How can we prove this series?
I try: can't think of any ideas
I just wonder if we can simplify ${(x-1)(x-2)\over (x+1)(x+2)}$ into a simple polynomial expression.
We can use Faling and rising factorials
$$-{C_{x-2}\over {2x-1\choose x}}=1-{(x)_2\over x^{(2)}}(2^2)+{(x)_3\over x^{(3)}}(3^2)-{(x)_4\over x^{(4)}}(4^2)+\cdots$$
$${(x)_n\over x^{(n)}}={{x\choose n}\over {x+n-1\choose n}}$$
$$-{C_{x-2}\over {2x-1\choose x}}=1-{{x\choose 2}\over {x+1\choose 2}}(2^2)+{{x\choose 3}\over {x+2\choose 3}}(3^2)-\cdots$$
From here I can't go further. Can anyone help me to prove this series?
A hint from @Thomas
$$-{x\over 2(2x-1)(2x-3)}=1-{{x\choose 2}\over {x+1\choose 2}}(2^2)+{{x\choose 3}\over {x+2\choose 3}}(3^2)-\cdots$$
As already suggested by @Thomas, the RHS can be transformed into one of these forms $$ \begin{gathered} - \frac{{C_{\,x - 2} }} {{\left( \begin{gathered} 2x - 1 \\ x \\ \end{gathered} \right)}} = - \frac{1} {{x - 1}}\left( \begin{gathered} 2\left( {x - 2} \right) \\ \left( {x - 2} \right) \\ \end{gathered} \right)/\left( \begin{gathered} 2x - 1 \\ x \\ \end{gathered} \right) = \hfill \\ = - \frac{{\left( {2\left( {x - 2} \right)} \right)^{\,\underline {\,x - 2\,} } x!}} {{\left( {x - 1} \right)\left( {x - 2} \right)!\left( {2x - 1} \right)^{\,\underline {\,x\,} } }} = - \frac{{\left( {2x - 4} \right)^{\,\underline {\,x - 2\,} } x}} {{\left( {2x - 1} \right)^{\,\underline {\,x\,} } }} = \hfill \\ = - \frac{{\left( {x - 1} \right)x}} {{\left( {2x - 1} \right)^{\,\underline {\,3\,} } }} = - \frac{{x^{\,\underline {\,2\,} } }} {{\left( {2x - 1} \right)^{\,\underline {\,3\,} } }} = - \frac{x} {{2\left( {2x - 1} \right)\left( {2x - 3} \right)}} = \hfill \\ = \frac{1} {{8\left( {2x - 1} \right)}} - \frac{3} {{8\left( {2x - 3} \right)}} \hfill \\ \end{gathered} \tag{1} $$
The RHS can instead be developed in one of these forms $$ \begin{gathered} f(x) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \left( {k + 1} \right)^{\,2} \frac{{x^{\,\underline {\,k + 1\,} } }} {{x^{\,\overline {\,k + 1\,} } }}} = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \left( {k + 1} \right)^{\,2} \left( {x - 1} \right)^{\,\underline {\, - \,\left( {k + 1} \right)\,} } x^{\,\underline {\,k + 1\,} } } = \hfill \\ = - \sum\limits_{0\, \leqslant \,k} {\frac{{\left( {k + 1} \right)^{\,2} x^{\,\underline {\,k + 1\,} } }} {{\left( { - x} \right)^{\,\underline {\,k + 1\,} } }}} = - \sum\limits_{0\, \leqslant \,k} {\left( {k + 1} \right)^{\,2} \frac{{\Gamma (x + 1)\Gamma ( - x - k)}} {{\Gamma (x - k)\Gamma ( - x + 1)}}} = - \frac{{\Gamma (x + 1)}} {{\Gamma ( - x + 1)}}\sum\limits_{0\, \leqslant \,k} {\left( {k + 1} \right)^{\,2} \frac{{\Gamma ( - x - k)}} {{\Gamma (x - k)}}} = \hfill \\ = \Gamma (x + 1)\Gamma (x)\sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\left( {k + 1} \right)^{\,2} }} {{\Gamma (x - k)\Gamma (x + k + 1)}}} \hfill \\ \end{gathered} $$ where the latter turns useful to check for convergence.
We will focalize our attention on the first form on the second line.
Let's premise that for $z,w,a \in C$ and within the definition domain of the Rising and Falling Factorials we have $$ \frac{{\left( {z + a} \right)^{\,\underline {\,w\,} } }} {{z^{\,\underline {\,w\,} } }}\, = \frac{{\left( {z + a} \right)^{\,\underline {\,a\,} } }} {{\left( {z + a - w} \right)^{\,\underline {\,a\,} } }}\quad \quad z^{\,\overline {\, - w\,} } = \frac{1} {{\left( {z - w} \right)^{\,\overline {\,w\,} } }} = \frac{1} {{\left( {z - 1} \right)^{\,\underline {\,w\,} } }} $$ by which we can transform the addenda, leaving out for the moment the $(k+1)^2$ factor, as
Now we have an expression where the summation index appears only in the base of the Falling Factorial and we know the Indefinite Summation (Anti-Delta) of such an expression, which is $$ \begin{gathered} \sum\nolimits_{\,z} {z^{\,\underline {\,w\,} } } = \Delta _{\,z} ^{\, - 1} \,z^{\,\underline {\,w\,} } = \frac{1} {{w + 1}}\,z^{\,\underline {\,w + 1\,} } + c \hfill \\ \sum\nolimits_{\,z} {\left( { - z} \right)^{\,\underline {\,w\,} } } = \Delta _{\,z} ^{\, - 1} \left( { - z} \right)^{\,\underline {\,w\,} } = - \frac{1} {{w + 1}}\left( { - z + 1} \right)^{\,\underline {\,w + 1\,} } + c \hfill \\ \sum\nolimits_{\;z\, = \,0}^{\;\infty } {\left( { - z + a} \right)^{\,\underline {\, - \,w\,} } } = \frac{1} {{1 - w}}\left( {a + 1} \right)^{\,\underline {\,1 - w\,} } \quad \left| {\;0 < \operatorname{Re} \left( w \right)} \right. \hfill \\ \end{gathered} \tag{3} $$ since it is $$ z^{\,\underline {\,w\,} } \, \approx \left| z \right|^{\,\operatorname{Re} (w)} \quad \left| {\;z\, \to \,\infty ,\;\;\left| {\,\arg (z)\,} \right| < \pi } \right. $$
Therefore what we have to do is to express $(k+1)^2$ in a form that can be aggregated to the factorial, i.e. $$ \left( {k + 1} \right)^{\,2} = \left( {x - x - k - 1} \right)^{\,2} = \left( { - x - k + 1} \right)^{\,\underline {\,2\,} } + \left( {2x - 3} \right)\left( { - x - k} \right)^{\,\underline {\,1\,} } + \left( {x - 1} \right)^{\,2} $$
So we can proceed with the final steps
which compared with (1) provides the Q.E.D.
--- Note ---
As per the comment by Markus, it shall be noted that, for $x \in \mathbb N$, the expressions given above shall be taken in the limit. Consider for ex. $x=2$ , then: $$ h(2,k) = \left( { - 1} \right)^{\,k} \frac{{2^{\,\underline {\,k + 1\,} } }} {{2^{\,\overline {\,k + 1\,} } }} = 1, - \frac{1} {3},0, \cdots \quad \left| {\;k = 0,1,2, \cdots } \right. $$ and $$ \begin{gathered} h(2,k) = - 2^{\,\underline {\,4\,} } \left( { - 2 - k - 1} \right)^{\,\underline {\, - \,4\,} } = - \frac{{2^{\,\underline {\,4\,} } }} {{\left( {2 - k - 1} \right)^{\,\underline {\,4\,} } }} = \hfill \\ = - \frac{{ - 2/\bar \infty }} {{2/\bar \infty }}, - \frac{{ - 2/\bar \infty }} {{ - 6/\bar \infty }}, - \frac{{ - 2/\bar \infty }} {{24}}, \cdots \hfill \\ \end{gathered} $$