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today,I see a amazing math problem:

show that


This problem is from here.

But I consider sometimes,and I think it maybe use Taylor therom


Thank you,Now is 24:00 in beijing time,so I must go bed.I hope someone can help.Thank you

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up vote 7 down vote accepted

The sum can be written as $$\sum_{n=0}^\infty \binom{4n}{n}\frac{1}{(3n+1)2^{4n+1}}=\frac{1}{2}{\;}_3F_2 \left(\frac{1}{4},\frac{1}{2},\frac{3}{4}; \frac{2}{3},\frac{4}{3};\frac{16}{27} \right)$$ For the Hypergeometric Function, see equation (25) of $${\;}_3F_2 \left(\frac{1}{4},\frac{1}{2},\frac{3}{4}; \frac{2}{3},\frac{4}{3};y \right) = \frac{1}{1-x^3}$$ where $x$ and $y$ are related by $$y=\left( \frac{4x(1-x^3)}{3}\sqrt[3]{4} \right)^3$$ Putting $y=16/27$, we get a polynomial equation in $x$. $$1=16 x^3(1-x^3)^3$$ According to Mathematica, this equation has only two real roots which are $$\begin{align*} \alpha &= \frac{1}{\sqrt[3]{2}} \\ \beta &= \left\{\frac{1}{6} \left(5-\frac{4}{\left(19-3 \sqrt{33}\right)^{1/3}}-\left(19-3 \sqrt{33}\right)^{1/3}\right) \right\}^{1/3} \end{align*}$$ Numericaly, $\beta$ gives the right answer. So, the final result is $$ \begin{align*}\sum_{n=0}^\infty \binom{4n}{n}\frac{1}{(3n+1)2^{4n+1}} &=\frac{1}{2\left(1-\beta^3 \right)} \\ &= \dfrac{\sqrt[3]{17+3\sqrt{33}}}{3}-\dfrac{2}{3\sqrt[3]{17+3\sqrt{33}}}-\dfrac{1}{3} \\ &\approx 0.543689\end{align*}$$

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The transformed series in Marin's post seems suspiciously like Taylor's expansion of Bring radical $$\begin{aligned}\sum_{n=1}^{\infty}\dfrac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)} &= \sum_{n=0}^{\infty} \binom{4n}{n} \frac{1}{(3n + 1)2^{4n + 1}} \\ &= \frac{16^{1/3}}{2} \sum_{n=0}^{\infty} \binom{4n}{n} \frac{\left ( 16^{-1/3} \right )^{3n + 1}}{(3n + 1)} \\ &= \frac{1}{2^{1/3}}\sum_{n = 0}^{\infty} \frac{\left(16^{-1/3}\right)^n}{n!} \left [\frac{d^{n-1}}{dz^{n-1}} \left ( 1 + z \right )^{4n} \right ]_{z = 0}\end{aligned}$$

The last sum is the Lagrange-Burmann inversion formula applied to $\phi(z)=(1+z)^4$. It is straightforward now that the series (exlcuding the constant factor $1/2^{1/3}$) is one of the two real roots of the quartic $z^4 + z + 16^{-1/3}$. One can exclude $-1/2^{1/3}$ via invoking a numerical check, and thus

$$\sum_{n=1}^{\infty}\dfrac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)} = \dfrac{\sqrt[3]{17+3\sqrt{33}}}{3}-\dfrac{2}{3\sqrt[3]{17+3\sqrt{33}}}-\dfrac{1}{3} \blacksquare$$

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Hint: the right side is a root of $27 z^3 + 18 z - 34$. There is a way to solve equations of the form $x^p + a x + b = 0$ using generalized hypergeometric functions.

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$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$

$\ds{% {\cal S} \equiv \sum_{n = 1}^{\infty}{{4n - 4 \choose n - 1} \over 2^{4n - 3}\,\pars{3n - 2}} = {1 \over 2} \sum_{n = 0}^{\infty}{4n \choose n}\,{\pars{1/16}^{n} \over 3n + 1} = {16^{1/3} \over 2} \sum_{n = 0}^{\infty}{4n \choose n}\,{\pars{16^{-1/3}}^{3n + 1} \over 3n + 1}}$.

$$ \mbox{Let's define}\quad {\rm f}\pars{z} \equiv \sum_{n = 0}^{\infty}{4n \choose n}\,{z^{3n + 1} \over 3n + 1} \quad\mbox{such that}\quad {\cal S} = {16^{1/3} \over 2}{\rm f}\pars{1 \over 16^{1/3}} $$

\begin{align} \!\!\!\!{\rm f}'\pars{z} &= \sum_{n = 0}^{\infty}z^{3n}{4n \choose n} = \sum_{n = 0}^{\infty}z^{3n}\int_{\verts{z'}\ =\ 1^{-}} {\pars{1 + z'}^{4n} \over z'^{n + 1}}\,{\dd z' \over 2\pi\ic} = \!\!\!\int_{\verts{z'}\ =\ 1^{-}}\!{\dd z' \over 2\pi\ic}\,{1 \over z'} \sum_{n = 0}^{\infty}\bracks{z^{3}\pars{1 + z'}^{4} \over z'}^{n} \\[3mm]&= \int_{\verts{z'}\ =\ 1^{-}}\!{\dd z' \over 2\pi\ic} {1 \over z' - z^{3}\pars{1 + z'}^{4}} \end{align}

With WA, we try to find the zeros of $ax^{4} - x + 1 = 0$ $\pars{~a \equiv z^{3}, \quad x \equiv z' + 1~}$ and they $\underline{\bf look}$ quite involved. The interesting value of $a$ is $a = z^{3} = 1/16$ and the behavior in $\pars{0,z_{m}}$ with $z_{m} > 16^{-1/3}$. With WA, we look for zeros of $x^{4} -16x + 16 = 0$. It $\underline{\bf yields}$ the following roots, in terms of $z'$: $$ z' = 1\,,\quad0.0874\,,\quad -2.5437 \pm 2.2303\,\ic $$ It's clear that there is one root close to zero but ${\bf \mbox{we don't know how the}\ z' = 1\ \mbox{root evolves when}\ z\ \mbox{moves away of}\ z = 16^{-1/3}}$. The contribution of the root $z_{0}$ "close to zero" is given by:

$$ \lim_{z' \to z_{0}}{z' - z_{0} \over z' - z^{3}\pars{1 + z'}^{4}} = {1 \over 1 - 4z^{3}\pars{1 + z_{0}}^{3}} = {1 \over 1 - 4z^{3}\pars{z_{0}/z^{3}}^{3/4}} = {1 \over 1 - 4z^{3/4}z_{0}^{3/4}} $$

Notice that $z_{0}$ is a function of $z$. We hope that a precise ( analytical ) solution of the zeros mentioned above will yields ${\rm f}\pars{z}$.

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