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The sum is: $$\frac12-\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10}-\frac{1\cdot3\cdot5\cdot7\cdot9\cdot11}{2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14}+-\dots$$ background: so the previous problem was the same question except that each minus sign was replaced with a plus sign, and for that case I managed to solve it by noticing that:$$(1-x)^{-1/2}=1+\frac{x}2+\frac{1\cdot3}{2\cdot4}x^2+\dots\text{ and }(1+x)^{1/2}=1-\frac{x}2+\frac{1\cdot3}{2\cdot4}x^2-\dots$$ Thus summing both sides after substituing $x^2$ for $x$ and multiplying in turn by $x$ one obtains:$$x+\frac{1\cdot3}{2\cdot4}x^5+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8}x^9+\dots=\frac12x[(1+x^2)^{-1/2}+(1-x^2)^{-1/2}]$$from which by integrating and putting $x=1$ one finally obtains the requested sum. The problem with the alternating one is that this simple trick is not possible here. What do you suggest? I should be able to do it without complex series (why I mention complex series? because it might be tempting to follow a similar reasoning identity for sum of binomial coefficients)

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According to Maple $$ \eqalign{\frac{1}{2} + \sum_{k=1}^\infty (-1)^k \frac{(4k-1)!!}{(4k+2)!!} & = \frac{1}{2} + \sum_{k=1}^\infty (-1)^k \frac{\Gamma(2k+1/2)}{2\sqrt{\pi} \Gamma(2k+2)}\cr &= \frac{i \sqrt{1-i}}{2} - \frac{i \sqrt{1+i}}{2} = \frac{\sqrt{2\sqrt{2}-2}}{2}}$$

EDIT: Hmm, where does this come from? If $a_k = (4k-1)!!/(4k+2)!!$, we have $$8 (k+1)(2k+3) a_{k+1} = (4k+1)(4k+3) a_k $$ This corresponds to the differential equation $$ -3 y + (24-32 x) y' + (16 x - 16 x^2) y'' = 0$$ which has solutions $\sqrt{1+\sqrt{x}}/\sqrt{x}$ and $\sqrt{1-\sqrt{x}}/\sqrt{x}$. And indeed for $$ g(x) = \frac{\sqrt{1+\sqrt{x}}}{2\sqrt{x}} - \frac{\sqrt{1-\sqrt{x}}}{2\sqrt{x}}$$ we have the Taylor series $$ \eqalign{g(x) &= \frac{1}{2} + \frac{x}{16} + \frac{7 x^2}{256} + \frac{33 x^3}{2048} + \ldots\cr &= 1 + \sum_{k=1}^\infty \frac{(4k-1)!!}{(4k+2)!!} x^k\cr}$$ and your series is the value for $x=-1$.

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In order to compute $$ S=\sum_{n\geq 0}\frac{(-1)^n}{4^{2n+1}}\binom{4n+2}{2n+1} $$ without resorting to complex numbers we may recall that $\frac{1}{4^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{\pi/2}\left(\sin\theta\right)^{2n}\,d\theta$, hence

$$ S = \frac{2}{\pi}\int_{0}^{\pi/2}\sum_{n\geq 0}(-1)^n\left(\sin\theta\right)^{4n+2}\,d\theta =\frac{2}{\pi}\int_{0}^{\pi/2}\frac{\sin^2\theta}{1+\sin^4\theta}\,d\theta=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{\cos^2\theta}{1+\cos^4\theta}\,d\theta$$ and by enforcing the substitution $\theta=\arctan u$ we are left with $$ S = \frac{2}{\pi}\int_{0}^{+\infty}\frac{dt}{1+(1+t^2)^2}$$ which can be computed through standard techniques.

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  • $\begingroup$ Awesome! Where can I learn this stuff? There is clearly some background knowledge here... $\endgroup$ – Peanut Jul 16 '20 at 17:11
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ The Sum is

$\ds{\bbox[15px,#ffd]{\frac12-\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10}-\frac{1\cdot3\cdot5\cdot7\cdot9\cdot11}{2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14}+-\dots}}:\ {\Large ?}$


\begin{align} &\bbox[15px,#ffe]{{1 \over 2} + \sum_{n = 1}^{\infty}\pars{-1}^{n}\ {\prod_{k = 1}^{2n}\pars{2k - 1} \over \prod_{q = 1}^{2n + 1}\pars{2q}}} = {1 \over 2} + \sum_{n = 1}^{\infty}\pars{-1}^{n}\ {2^{2n}\prod_{k = 1}^{2n}\pars{k - 1/2} \over 2^{2n +1}\prod_{q = 1}^{2n + 1}q} \\[5mm] = &\ {1 \over 2} + {1 \over 2}\sum_{n = 1}^{\infty}\pars{-1}^{n}\ {\pars{1/2}^{\overline{2n}} \over \pars{2n + 1}!} = {1 \over 2} + {1 \over 2}\sum_{n = 1}^{\infty}\pars{-1}^{n}\ {\Gamma\pars{1/2 + 2n}/\Gamma\pars{1 /2} \over \pars{2n + 1}!} \\[5mm] = &\ {1 \over 2} + {1 \over 2}\,{\pars{-3/2}! \over \pars{-1/2}!} \sum_{n = 1}^{\infty}\pars{-1}^{n}\ {\pars{2n - 1/2}! \over \pars{2n + 1}!\pars{-3/2}!} \\[5mm] = &\ {1 \over 2} - \sum_{n = 1}^{\infty}\pars{-1}^{n}{2n - 1/2 \choose 2n + 1} \\[5mm] = &\ {1 \over 2} - \sum_{n = 1}^{\infty}\pars{-1}^{n} {1/2 \choose 2n + 1}\pars{-1}^{2n + 1} = {1 \over 2} - \ic\sum_{n = 1}^{\infty}\ic^{2n + 1} {1/2 \choose 2n + 1} \\[5mm] = &\ {1 \over 2} - \ic\sum_{n = 3}^{\infty}\ic^{n} {1/2 \choose n}\,{1 - \pars{-1}^{n} \over 2} = {1 \over 2} + \Im\sum_{n = 3}^{\infty}{1/2 \choose n}\ic^{n} \\[5mm] = &\ {1 \over 2} + \Im\bracks{\pars{1 + \ic}^{1/2} - {1/2 \choose 0}\ic^{0} - {1/2 \choose 1}\ic^{1} - {1/2 \choose 2}\ic^{2}} = \Im\pars{1 + \ic}^{1/2} \\[5mm] = &\ \Im\pars{\root{2}\expo{\ic\pi/4}}^{1/2} = 2^{1/4}\Im\expo{\ic\pi/8} = \bbx{2^{1/4}\sin\pars{\pi \over 8}} \\[5mm] = &\ \bbx{2^{-3/4}\root{2 - \root{2}}}\ \approx 0.4551 \end{align}
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  • $\begingroup$ It seems that you can't do without complex numbers... $\endgroup$ – Peanut Jul 16 '20 at 16:30
  • $\begingroup$ @Dude I guess so$\ldots$ $\displaystyle\left\{\bullet \quad \bullet \atop {\mid \atop {\LARGE\smile}}\right\}$ $\endgroup$ – Felix Marin Jul 18 '20 at 18:39

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