How to prove double factorial sum: $\pi\frac{(-1)^{n+1}}{2^{2n-3}}\sum_{k=0}^{n-1}(-1)^k{2n\choose k}(n-k)=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}$?

Question

$$\pi\cdot\frac{(-1)^{n+1}}{2^{2n-3}}\sum_{k=0}^{n-1}(-1)^k{2n\choose k}(n-k)=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2},\ \ \ \ n\in\Bbb{N}^+\tag1$$

I have a feeling that $(1)$ might be true. Contour integration yielded the LHS while the RHS is given in the answer key. To prove this result without induction, I first I distributed the sum, and showed that

$$\begin{align} n\sum_{k=0}^{n-1}(-1)^k{2n\choose k}&=n\sum_{k=0}^{n-1}(-1)^k{2n\choose k}{n-k-1\choose n-k-1}\\&=n\sum_{k=0}^{n-1}(-1)^{k+n-k-1}{2n\choose k}{-1\choose n-k-1}\\ &=n(-1)^{n-1}{2n-1\choose n-1}.\end{align}$$

I am not sure what to do with

$$\sum_{k=1}^{n-1}k(-1)^k{2n\choose k}\tag2$$

since this has the additional $k$. I thought about using some definite integral substitutions for $k$ but those did not get me anywhere. A hint to solving $(2)$ ought to suffice though complete answers are also welcome.

EDIT. The original equality in $(1)$ does not hold. I had messed up the denominator. The correct identity is

$$\pi\cdot\frac{(-1)^{n+1}}{2^{2n-1}}\sum_{k=0}^{n-1}(-1)^k{2n\choose k}(n-k)=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2},\ \ \ \ n\in\Bbb{N}^+\tag{1'}.$$

Thanks to Felix Marin for pointing this out!

Answer 1

We have $$ k\binom{2n}{k} = \frac{(2n)!}{(k-1)!(2n-k)!} = 2n\binom{2n-1}{k-1} $$

hence it follows that:

$$ A(n) = \sum_{k=1}^{n-1} k(-1)^k \binom{2n}{k} = 2n\sum_{k=1}^{n-1}(-1)^k \binom{2n-1}{k-1} = -2n\sum_{k=0}^{n-2}(-1)^k \binom{2n-1}{k} $$ and you already know how to deal with the last sum by exploiting symmetry.

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An alternative comes from experience: the RHS of your identity is clearly related with integrals of the form $\int_{0}^{\pi/2}\sin(x)^{2n}\,dx$. So we may start considering that

$$(e^{it}-e^{-it})^{2n} = \sum_{k=0}^{2n}\binom{2n}{k}e^{(2n-k)it}e^{-kit}(-1)^k $$ differentiate both sides with respect to $t$ $$ \frac{d}{dt}\left(2i\sin t\right)^{2n} = 2i\sum_{k=0}^{2n}\binom{2n}{k}(-1)^k(n-k)e^{(2n-2k)it} $$ then isolate the contributes given by $k\in\{0,\ldots,n-1\}$ by multiplying both sides by $(e^{-2nit}+e^{-(2n-2)it}+\ldots+e^{-2it})$ (that is a simple geometric sum) and integrating over $(0,2\pi)$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\pi\,{\pars{-1}^{n + 1} \over 2^{2n - 3}} \sum_{k = 0}^{n - 1}\pars{-1}^{k}{2n \choose k}\pars{n - k} = \require{cancel}\cancel{{\pars{2n - 3}!! \over \pars{2n - 2}!!}\,{\pi \over 2}}:\ {\large ?}\,, \qquad n \in \mathbb{N}_{\ >\ 0}}$.

$$ \bbox[15px,#ffe,border:1px dotted navy]{\ds{% \mbox{The RHS}\ correct\ answer\ \mbox{is}\quad {\pi \over 2^{2n - 3}}{2n - 2 \choose n - 1}}} $$ \begin{align} &\sum_{k = 0}^{n - 1}\pars{-1}^{k}{2n \choose k}\pars{n - k} = \sum_{k = 0}^{n - 1}\pars{-1}^{k}\pars{n - k}{2n \choose 2n - k} \\[5mm] = &\ \sum_{k = 0}^{n - 1}\pars{-1}^{k}\pars{n - k}\ \overbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{2n} \over z^{2n - k + 1}} \,{\dd z \over 2\pi\ic}}^{\ds{2n \choose 2n - k}} \\[5mm] = &\ \oint_{\verts{z} = 1}{\pars{1 + z}^{2n} \over z^{2n + 1}} \overbrace{\sum_{k = 0}^{n - 1}\pars{-1}^{k}\pars{n - k}z^{k}} ^{\ds{{n \over 1 + z} + {z \over \pars{1 + z}^{2}} + \pars{-1}^{n + 1}\,{z^{n + 1} \over \pars{1 + z}^{2}}}}\ \,{\dd z \over 2\pi\ic} \\[5mm] = &\ n\ \underbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{2n - 1} \over z^{2n + 1}} \,{\dd z \over 2\pi\ic}}_{\ds{{2n - 1 \choose 2n}\ =\ 0}}\ +\ \underbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{2n - 2} \over z^{2n}} \,{\dd z \over 2\pi\ic}}_{\ds{{2n - 2 \choose 2n - 1}\ =\ 0}}\ +\ \pars{-1}^{n + 1}\oint_{\verts{z} = 1}{\pars{1 + z}^{2n - 2} \over z^{n}} \,{\dd z \over 2\pi\ic} \\[5mm] = &\ \pars{-1}^{n + 1}{2n - 2 \choose n - 1} \end{align}

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$$ \bbx{\ds{\pi\,{\pars{-1}^{n + 1} \over 2^{2n - 3}} \sum_{k = 0}^{n - 1}\pars{-1}^{k}{2n \choose k}\pars{n - k} = {\pi \over 2^{2n - 3}}{2n - 2 \choose n - 1}}} $$