Prove $_4 F_3\left(2,\frac32,\frac32,\frac32;\frac52,\frac52,\frac52;1\right)=\frac{27}{16}\left(\pi^2-7\zeta(3)\right) $

Question

How to prove the following result about the generalized hypergeometric function $_4 F_3$? $$_4 F_3\left(2,\frac32,\frac32,\frac32;\frac52,\frac52,\frac52;1\right)\stackrel{?}=\frac{27}{16}\left(\pi^2-7\zeta(3)\right) $$ Apart from the definition, I know very little about hypergeometric functions. $$_4 F_3(a_1,a_2,a_3,a_4;b_1,b_2,b_3;z)=\sum_{n=0}^\infty \frac{(a_1)_n(a_2)_n(a_3)_n(a_4)_n}{(b_1)_n(b_2)_n(b_3)_n}\frac{z^n}{n!} $$ This particular $_4 F_3$ is not in one of the many known closed forms provided by Wolfram. I also tried getting an idea by reading Jack D'aurizio's paper Surprising identities for the hypergeometric $_4 F_3$ function, but I couldn't see how to attack this series.

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Could you please suggest any ideas on how to prove it?

Answer 1

Simplify the Pochhammer symbols with the identity $(a)_n=\dfrac{\Gamma(a+n)}{\Gamma(a)}$ and you'll end up with zeta-related sums:

$$\begin{align*} {}_4F_3\left(\left.\begin{array}{c}2,\frac32,\frac32,\frac32\\\frac52,\frac52,\frac52\end{array} \right\vert 1\right) &= \sum_{n=0}^\infty \frac{(2)_n \left[\left(\frac32\right)_n\right]^3}{n! \left[\left(\frac52\right)_n\right]^3} \\ &= \sum_{n=0}^\infty \frac{\frac{\Gamma(n+2)}{\Gamma(2)} \left(\frac{\Gamma\left(n+\frac32\right)}{\Gamma\left(\frac32\right)}\right)^3}{\Gamma(n+1) \left(\frac{\Gamma\left(n+\frac52\right)}{\Gamma\left(\frac52\right)}\right)^3} \\ &= \left(\frac32\right)^3 \sum_{n=0}^\infty \frac{n+1 }{\left(n+\frac32\right)^3} \\ &= 27 \sum_{n=1}^\infty \frac{n}{(2n+1)^3} \\ &= \frac{27}2 \sum_{n=1}^\infty \left(\frac1{(2n+1)^2} - \frac1{(2n+1)^3}\right) \end{align*}$$

Now observe

$$\begin{align*} \zeta(a) &= \sum_{n=1}^\infty \frac1{n^a} \\ &= 1 + \sum_{n=1}^\infty \frac1{(2n)^a} + \sum_{n=1}^\infty \frac1{(2n+1)^a} \\[1ex] \implies \sum_{n=1}^\infty \frac1{(2n+1)^a} &= \left(1-2^{-a}\right)\zeta(a)-1 \end{align*}$$

Answer 2

The following is a proof of the identity $$\small{}_4F_3\left(\left.\begin{array}{c}b,a,a,a\\a+1,a+1,a+1\end{array} \right\vert 1\right) = \frac{a^{3}}{2} B(a, 1-b) \left[\left(\psi(a) -\psi(a-b+1) \right)^{2}+ \psi^{(1)}(a)- \psi^{(1)}(a-b+1) \right],$$ for $\Re(a) >0$ and $ \Re(b) < 3. $

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For $\Re(x) > 0$ and $\Re(y) > -2$, the second derivative of the beta function $B(x,y)$ with respect to $x$ has the integral representation $$\frac{\partial^{2} }{\partial x^{2}} \, B(x,y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} \ln^{2}(t)\, \mathrm dt.$$

Expanding $(1-t)^{y-1}$ in a Taylor series at $t=0$ and switching the order of summation and integration, as was done in the answer to this question, we get $$ \begin{align} \frac{\partial^{2} }{\partial x^{2}} \, B(x,y) &= \sum_{n=0}^{\infty} (-1)^{n}\binom{y-1}{n} \int_{0}^{1} t^{x+n-1} \ln^{2}(t) \, \mathrm dt \\ &= \sum_{n=0}^{\infty} (-1)^{n}\binom{y-1}{n} \frac{2}{(x+n)^{3}} \\ &= 2 \sum_{n=0}^{\infty} \frac{(1-y)_{n}}{(x+n)^{3}n!} . \end{align}$$

So for $\Re(b) <3$, we have$$ \begin{align} _4 F_3\left(b,a,a,a;a+1,a+1,a+1;1\right) &= \sum_{n=0}^{\infty} \frac{(b)_{n} \, \left((a)_{n}\right)^{3}}{\left((a+1)_{n}\right)^{3}} \frac{1}{n!} \\ &= a^{3}\sum_{n=0}^{\infty} \frac{(b)_{n}}{(a+n)^{3}\, n!} \\ &= \frac{a^{3}}{2} \frac{\partial^{2} }{\partial a^{2}} \, B(a,1-b). \end{align}$$

Differentiating the defintion $B(a,1-b)= \frac{\Gamma(a) \Gamma(1-b)}{\Gamma(a-b+1)}$ with respect to $a$, we get $$ \begin{align} \frac{\partial }{\partial a} B(a,1-b) &= \frac{\partial }{\partial a} \frac{\Gamma(a) \Gamma(1-b)}{\Gamma(a-b+1)} \\ &= \Gamma(1-b) \, \frac{\Gamma'(a)\Gamma(a-b+1)-\Gamma(a)\frac{\partial }{\partial a} \Gamma(a-b +1)}{\Gamma^{2}(a-b+1)} \\ &= \frac{\Gamma(1-b)}{\Gamma(a-b+1)} \left(\Gamma'(a)-\Gamma(a) \, \psi(a-b+1) \right) \\ &= \frac{\Gamma(a)\Gamma(1-b)}{\Gamma(a-b+1)} \left(\psi(a)- \psi(a-b+1) \right) \\ &= B(a, 1-b)\left(\psi(a)- \psi(a-b+1) \right). \end{align}$$

And differentiating with respect to $a$ a second time, we get

$\begin{align} \frac{\partial^{2} }{\partial a^{2}} B(a, 1-b) &= \left(\psi(a)- \psi(a-b+1) \right) \frac{\partial }{\partial a}B(a, b-1) + B(a, 1-b) \frac{\partial }{\partial a}\left(\psi(a)- \psi(a-b+1) \right) \\ &= B(a, 1-b) \left(\psi(a)- \psi(a-b+1) \right)^{2}+ B(a,1-b) \left(\psi^{(1)}(a) - \psi^{(1)}(a-b+1) \right) \\ &= B(a,1-b) \left(\left(\psi(a)- \psi(a-b+1) \right)^{2} + \psi^{(1)}(a) - \psi^{(1)}(a-b+1) \right). \end{align}$

Therefore, $$\small {}_4F_3\left(\left.\begin{array}{c}b,a,a,a\\a+1,a+1,a+1\end{array} \right\vert 1\right) = \frac{a^{3}}{2} B(a, 1-b) \left[\left(\psi(a) -\psi(a-b+1) \right)^{2}+ \psi^{(1)}(a)- \psi^{(1)}(a-b+1) \right] $$ for $\Re(a) >0$ and $\Re(b) <3$.

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Call the function on the right side of the above equation $f(a,b)$.

Since it's not particularly obvious, let's check that the limits $\lim_{b \to 1} f(a,b)$ and $\lim_{b \to 2} f(a,b)$ exist and are finite.

Expanding at $b=1$, we get $$\begin{align} \lim_{b \to 1} f(a,b) &= \small \frac{a^{3}}{2} \lim_{b \to 1} \left(-\frac{1}{b-1} + O(1) \right) \left((b-1)^{2}\psi^{(1)}(a)^{2} +O\left((b-1)^{3}\right)+(b-1)\psi^{2}(a) +O\left((b-1)^{2} \right)\right) \\ & = - \frac{a^{3}}{2} \, \psi^{(2)}(a). \end{align}$$

And if $a \ne 1$,

$ \begin{align} \lim_{b \to 2} f(a,b) = \frac{a^{3}}{2} (a-1) \lim_{b \to 2} & \left(\frac{1}{b-2} + O(1) \right) \Big(\frac{1}{(a-1)^{2}}+ \frac{2(b-2)}{a-1} \psi^{(1)}(a-1) + O\left((b-2)^{2} \right) \\ &- \frac{1}{(a-1)^{2}}+(b-2) \psi^{(2)}(a-1) + O\left((b-2)^{2} \right)\Big) \\ &= \frac{a^{3}}{2} \left( 2 \psi^{(1)}(a-1) +(a-1) \psi^{(2)}(a-1) \right) \\ &= \frac{a^{3}}{2} \left( 2 \left(\psi^{(1)}(a) + \frac{1}{(a-1)^{2}} \right) +(a-1) \left(\psi^{(2)}(a) - \frac{2}{(a-1)^{3}} \right) \right) \\ &=\frac{a^{3}}{2} \left( 2 \psi^{(1)}(a) +(a-1) \psi^{(2)}(a) \right). \end{align}$

A quick check confirms that $\sum_{n=0}^{\infty} \frac{(2)_{n}}{(1+n)^{3}n!} = \psi^{(1)}(1) = \zeta(2).$