Generalized hypergeometric functions X

Series involving the binomial coefficient

Today we show the proof of this series proposed by @qq3023625451 \[ \sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)}{2^{10n}}{\binom{2n}{n}}^5 = \frac{2}{\Gamma^{4}\left(\frac{3}{4}\right)}\] In the proof we use some properties of the generalized hypergeometric function ${}_pF_{q}$

Proof:

The Pocchhammer symbol (rising factorial) is defined as the product: \[ (x)_{n} = x(x+1)\cdots(x+n-1) = \prod_{j=0}^{n-1}(x+j) \] while the generalized hypergeometric function is defined \[ {}_pF_{q} \left[{a_{1},a_2,...,a_{p}\atop b_{1},b_{2},...,b_{q}}; z \right] = \sum_{n=0}^{\infty} \frac{(a_{1})_{n}(a_{2})_{n}\cdots (a_{p})_n}{(b_{1})_{n}(b_{2})_{n}\cdots (b_{q})_{n} } \frac{z^n}{n!} \]

Note that $(1)_{n} = n!$

The rising factorial obey the duplication formula: \[(x)_{2n} = 2^{2n} \left(\frac{x}{2}\right)_{n} \left(\frac{1+x}{2}\right)_{n} \tag{1} \] and the recursion formula for the argument \[ (n+x)(x)_{n} = x(x+1)_{n} \tag{2} \] Hence \begin{align*} S = \sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)}{2^{10n}}{\binom{2n}{n}}^5 =& \sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)(2n)!^5}{ 2^{10n}n!^{10}}\\ =& \sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)(1)_{2n}^5}{ 2^{10n}(1)_{n}^{10}}\\ =& \sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)2^{10n}\left(\frac{1}{2}\right)_{n}^{5}(1)_{n}^{5}}{ 2^{10n}(1)_{n}^{10}}\quad \textrm{from (1)}\\ =& \sum_{n=0}^{\infty} \frac{(-1)^n4\left(n+\frac{1}{4}\right)\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n}{ (1)_n(1)_n(1)_n(1)_nn!}\\ =& \sum_{n=0}^{\infty} \frac{\left(\frac{5}{4}\right)_{n}\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n}{\left(\frac{1}{4}\right)_{n} (1)_n(1)_n(1)_n(1)_n} \frac{(-1)^n}{n!} \quad \textrm{from (2)}\\ =& {}_{6}F_{5} \left[ {\frac{1}{2}, \frac{5}{4}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \atop \frac{1}{4},1,1,1,1};-1 \right] \end{align*} One of the Whipple's identities states that \[ {}_{6}F_{5} \left[{a, 1+\frac{1}{2}a , b,c,d,e \atop \frac{1}{2}a, 1+a-b, 1+a-c, 1+a-d, 1+a-e }; -1\right] = \frac{\Gamma(1-a-d)\Gamma(1+a-e)}{\Gamma(1+a)\Gamma(1+a-d-e)} {}_{3}F_{2}\left[ {1+a-b-c,d,e \atop 1+a-b,1+a-c}; 1\right]\] Hence \[ S = {}_{6}F_{5} \left[ {\frac{1}{2}, \frac{5}{4}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \atop \frac{1}{4},1,1,1,1};-1 \right]= \frac{1}{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{1}{2}\right)} {}_{3}F_{2} \left[{\frac{1}{2},\frac{1}{2},\frac{1}{2} \atop 1, 1}; 1\right]\] ${}_{3}F_{2}$ obey the Dixon's well poised sum: \[{{}_{3}F_{2}}\left[{a,b,c\atop a-b+1,a-c+1};1\right]=\frac{\Gamma\left(\frac{1% }{2}a+1\right)\Gamma\left(a-b+1\right)\Gamma\left(a-c+1\right)\Gamma\left(% \frac{1}{2}a-b-c+1\right)}{\Gamma\left(a+1\right)\Gamma\left(\frac{1}{2}a-b+1% \right)\Gamma\left(\frac{1}{2}a-c+1\right)\Gamma\left(a-b-c+1\right)},\] Hence, using $\displaystyle \Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{2}$ and $\displaystyle \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi} $ \begin{align*} S = \frac{1}{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{1}{2}\right)} {}_{3}F_{2} \left[{\frac{1}{2},\frac{1}{2},\frac{1}{2} \atop 1, 1}; 1\right] =& \frac{\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{1}{2}\right)} \\ =& \frac{4\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{1}{4}\right)}{\pi^2 \Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)} \\ =& \frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{1}{4}\right)}{\pi^2 \Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)} \quad \left(\Gamma(z+1) = z\Gamma(z) \right) \\ =& \frac{2}{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)} \quad \left( \Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin\left(\pi z\right)}\right)\\ =& \frac{2}{\Gamma^{4}\left(\frac{3}{4}\right)} \end{align*} \[ \boxed{\sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)}{2^{10n}}{\binom{2n}{n}}^5 = \frac{2}{\Gamma^{4}\left(\frac{3}{4}\right)}}\]