Prove $\,_3F_2\left ( \frac12,\frac12,1;\frac34,\frac34;-\frac18 \right ) =\frac23+\frac{\Gamma\left ( \frac34 \right )^2}{3\sqrt{\pi} }$

Question

With some experiments, I discover an identity, which eventually evaluates to $$ \,_3F_2\left ( \frac12,\frac12,1;\frac34,\frac34;-\frac18 \right ) =\frac23+\frac{\Gamma\left ( \frac34 \right )^2}{3\sqrt{\pi} }. $$ This is special for the factor $-1/8$ and its simple result. I hope that there exist some ways we could attack it, and thank for your creative efforts.

Answer 1

One can give a very short and elementary proof using WZ-pairs.

We have the following analytic fact: if $F,G$ satisfy $$\tag{WZ}F(n+1,k) - F(n,k) = G(n,k+1)-G(n,k)$$ and moreover

• $\lim_{n\to\infty} G(n,k) = 0$ for each $k\geq 0$
• $\sum_{k\geq 0} F(0,k)$ converges
• $\sum_{n\geq 0} G(n,0)$ converges

then $\lim_{n\to\infty} \sum_{k\geq 0} F(n,k)$ exists and is finite, also $$\tag{1}\sum_{k\geq 0} F(0,k) = \sum_{n\geq 0} G(n,0) + \lim_{n\to\infty} \sum_{k\geq 0} F(n,k)$$

Proof sketch: We apply $\sum_{n= 0}^{N-1} \sum_{k=0}^{K-1}$ on both sides of $(\text{WZ})$, giving $$\sum_{k=0}^{K-1} (F(N,k)-F(0,k)) = \sum_{n= 0}^{N-1} (G(n,K) - G(n,0))$$ by our assumption, we can let $K\to \infty$ while fixing $N$, to obtain $$\sum_{k\geq 0} F(N,k) - \sum_{k\geq 0} F(0,k) = -\sum_{n= 0}^{N-1} G(n,0)$$ then letting $N\to\infty$ proves the claim.

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Now let $$F(n,k) = \frac{2^{-2 k-3 n-2} \Gamma (n+1) \Gamma \left(2 k+n-\frac{1}{2}\right)}{\Gamma \left(-n-\frac{1}{2}\right) \Gamma \left(n+\frac{3}{4}\right) \Gamma \left(k+n+\frac{3}{4}\right) \Gamma (k+n+1)}\quad G(n,k) = \frac{2(2 k+3 n+1)}{4 n+3}F(n,k)$$ one checks $(\text{WZ})$ is satisfied. All three bullets are also met, and $\lim_{n\to\infty} \sum_{k\geq 0} F(n,k) = 0$. Therefore we have $$\sum_{k\geq 0} F(0,k) = \sum_{n\geq 0} G(n,0)$$ explicitly: $$\sum_{k\geq 0} \frac{2^{-2 k-1} \left(\frac{1}{2}\right)_{2 k}}{(4 k-1) \left(\frac{3}{4}\right)_k (1)_k} = \sum_{n\geq 0} \frac{(-1/8)^n (2 n+1) (3 n+1) (1/2)_n^2}{(2 n-1) (4 n+3) (3/4)_n^2}$$

LHS can be summed using Gauss's $_2F_1$ theorem, giving $-\frac{\Gamma(3/4)^2}{2 \sqrt{\pi }}$. The summand of RHS is

$$-\Delta_n\left(\frac{(-1/8)^n(4 n-1) (1/2)_n^2}{(2 n-1) (3/4)_n^2}\right)-\frac{3}{2}\frac{(-1/8)^n (1/2)_n^2}{(3/4)_n^2}$$

here $\Delta_n f(n) = f(n+1)-f(n)$. From this we can get the value of $\sum_{n\geq 0} \frac{(-1/8)^n (1/2)_n^2}{(3/4)_n^2}$, which is OP's $_3F_2$ series.

Answer 2

Another solution is added for details. And this is also the "solution" that how I came up with the Hypergeometric series. At the beginning, it's more than easy to check the branch property of $K(k)$ defined by $\int_{0}^{1} \frac{1}{\sqrt{1-t^2}\sqrt{1-k^2t^2} }\text{d}t$, since it resembles functions like $(1-x^2)^{-1/2}$ . We have $$ K(k_{\pm})=\frac1k\left [ K\left ( \frac1k \right ) \pm iK^\prime\left ( \frac1k \right ) \right ] $$ when $\Re(k)>1.$ The simple formula leads to great achievements. By contour integrations, millions of new integrals are produced. Choosing $f(z)=\frac{K(z)}{\left ( 1+z^2 \right ) \sqrt{1-z^2} }$ and after some clear manipulations we could get $$ \int_{0}^{1} \frac{K(k)+K^\prime(k)}{\left ( 1+k^2 \right ) \sqrt{1-k^2} } \text{d}k=\frac{\Gamma\left ( \frac14 \right )^2 }{16} \left ( \frac{\Gamma\left ( \frac14 \right )^2 }{\pi}+\sqrt{\pi} \right ). $$ Then we do the following steps: \begin{align} \int_{0}^{1} \frac{K(k)+K^\prime(k)}{\left ( 1+k^2 \right ) \sqrt{1-k^2} } \text{d}k&=\frac{\Gamma\left ( \frac14 \right )^2 }{16} \left ( \frac{\Gamma\left ( \frac14 \right )^2 }{\pi}+\sqrt{\pi} \right )\\ &=\int_{0}^{1} \frac{K(k)}{\left ( 1+k^2 \right ) \sqrt{1-k^2} } \text{d}k+\int_{0}^{1} \frac{K^\prime(k)}{\left ( 1+k^2 \right ) \sqrt{1-k^2} }\text{d}k\\ &=\int_{0}^{1} \frac{K(k)}{\left ( 1+k^2 \right ) \sqrt{1-k^2} } \text{d}k+\int_{0}^{1} \frac{K(k)}{\left ( 2-k^2 \right ) \sqrt{1-k^2} }\text{d}k\\ &=3\int_{0}^{1} \frac{K(k)}{\left ( 1+k^2 \right )\left ( 2-k^2 \right ) \sqrt{1-k^2} } \text{d}k\\ &=\frac34\int_{0}^{1} \frac{K\left ( \sqrt{x} \right ) }{\left ( 1+\frac{x}2(1-x) \right ) \sqrt{x\left ( 1-x \right ) } } \text{d}x\\ &=\frac34\sum_{n\ge0}\frac{(-1)^n}{2^n} \int_{0}^{1} x^{n-\frac12}(1-x)^{n-\frac12}K\left ( \sqrt{x} \right ) \text{d}x\\ &=\frac34\sum_{n\ge0}\frac{(-1)^n}{2^n} \frac{\Gamma\left ( \frac14 \right )^2}{4\cdot2^{2n}} \frac{\Gamma\left ( n+\frac12 \right )^2 }{\Gamma\left ( n+\frac34 \right )^2 }\\ &=\frac{3\,\Gamma\left ( \frac14 \right )^4}{32\pi} \,_3F_2\left ( \frac12,\frac12,1;\frac34,\frac34;-\frac18 \right ). \end{align} Therefore by these equations we conclude that $$ \,_3F_2\left ( \frac12,\frac12,1;\frac34,\frac34;-\frac18 \right ) =\frac23+\frac{\Gamma\left ( \frac34 \right )^2 }{3\sqrt{\pi} }. $$ The process gives me some ideas, as stated following:

1. Our solution is highly relied on the unique symmetry of $K(k)$. It seems that generalizations are possible to certain $\,_2F_1$ but still there are some work needed doing.
2. Both the evaluation and result are surprisingly elegant in the form. I believe that some inside natures are going to be unearthed but so far effective efforts are not enough, and still confuse me.