Ratio of 2 Sums of products of binomial coefficients

Question

I want to prove that for $k \ even, 0 \leq k<n, n\in \mathbb{N}$:

$-\frac{1}{(2n-3-k)(k+2)}\sum \limits_{i=0}^{k} \frac{(-1)^{i} 2^{i} (2n-2-i)!}{(n-1-i)!i!(k-i)!}=\sum \limits_{i=0}^{k+2} \frac{(-1)^{i} 2^{i} (2n-2-i)!}{(n-1-i)!i!(k+2-i)!}$ (1).

I tried similar things like induction over $k$ or saying sth useful about $\frac{(-1)^{i} 2^{i} (2n-2-i)!}{(n-1-i)!i!(k-i)!}$.

When I compute $\sum \limits_{i=0}^{k} \frac{(-1)^{i} 2^{i} (2n-2-i)!}{(n-1-i)!i!(k-i)!}$ in Maple, I get $\sum \limits_{i=0}^{k} \frac{(-1)^{i} 2^{i} (2n-2-i)!}{(n-1-i)!i!(k-i)!}=\frac{- \pi (-1)^{\frac{1}{2}k+n}2^{2n}}{4 \Gamma(\frac{1}{2}k-n+\frac{3}{2})\Gamma(k+1) \Gamma(\frac{1}{2}-\frac{1}{2}k)}$ (2).

With (2) it is not too complicated to prove (1).

Are there any ideas how to tackle (1) or (2)? Thank you.

Answer 1

To tackle $(2)$:

$$\begin{eqnarray*}\sum_{i=0}^{k}\frac{(-1)^i 2^i (2n-2-i)!}{(n-1-i)!i!(k-i)!}&=&\frac{1}{k!}\sum_{i=0}^{k}\binom{k}{i}\frac{(-1)^i 2^i(2n-2-i)!}{(n-1-i)!}\\&=&\frac{(n-1)!}{k!}\sum_{i=0}^{k}\binom{k}{i}(-1)^i 2^i B(2n-2-i,n-1)\\&=&\frac{(n-1)!}{k!}\sum_{i=0}^{k}\binom{k}{i}(-2)^i \int_{0}^{1}(1-x)^n x^{2n-1-i}\,dx\\&=&\frac{(n-1)!}{k!}\int_{0}^{1}(1-x)^n x^{2n-1}\left(\frac{x-2}{x}\right)^k\,dx.\end{eqnarray*}$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{#c00000}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \sum_{i\ =\ 0}^{k}{\pars{-1}^{i}2^{i}\pars{2n - 2 - i}!\over \pars{n - 1 - i}!\, i!\, \pars{k - i}!}} ={\pars{n - 1}! \over k!} \color{#c00000}{\sum_{j\ =\ 0}^{k}{2n - 2 - j \choose n - 1 - j} {k \choose j}\pars{-2}^{j}} \end{align}

\begin{align}&\color{#c00000}{% \sum_{j\ =\ 0}^{k}{2n - 2 - j \choose n - 1 - j}{k \choose j}\pars{-2}^{j}} =\sum_{j\ =\ 0}^{k}{k \choose j}\pars{-2}^{j}\ \overbrace{% \oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n - 2 - j} \over z^{n - j}} \,{\dd z \over 2\pi\ic}}^{\ds{\color{#c00000}{{2n - 2 - j \choose n - 1 - j}}}} \\[5mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n - 2} \over z^{n}} \sum_{j\ =\ 0}^{k}{k \choose j}\pars{-\,{2z \over 1 + z}}^{j} \,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n - 2} \over z^{n}} \bracks{1 + \pars{-\,{2z \over 1 + z}}}^{k}\,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ 1} {\pars{1 + z}^{2n - 2 - k}\pars{1 - z}^{k} \over z^{n}} \,{\dd z \over 2\pi\ic} \\[5mm]&={1 \over \pars{n - 1}!}\,\lim_{z\ \to\ 0} \partiald[n - 1]{\bracks{\pars{1 + z}^{2n - 2 - k}\pars{1 - z}^{k}}}{z} \end{align}

Then, \begin{align} &\color{#66f}{\large% \sum_{i\ =\ 0}^{k}{\pars{-1}^{i}2^{i}\pars{2n - 2 - i}!\over \pars{n - 1 - i}!\, i!\, \pars{k - i}!} ={1 \over k!}\,\lim_{z\ \to\ 0} \partiald[n - 1]{\bracks{\pars{1 + z}^{2n - 2 - k}\pars{1 - z}^{k}}}{z}} \end{align}