Generalized hypergeometric functions VII

Nice series involving the harmonic number $H_{2n}$

Totday we show the proof of this nice series posted by @BenriBot_ccbs \[ \sum_{n=1}^{\infty} \frac{(-1)^nH_{2n}}{2n+1} = -\frac{\pi\ln(2)}{8}\] The proof relies on some properties of the rising factorial and the Gauss hypergometric function ${}_{2}F_{1}$.

Proof

Consider the following series: \[ S= \sum_{n=0}^{\infty} \frac{(t)_{2n}}{(2)_{2n}} x^n \] where \[(x)_{n} = x(1+x)\cdots (x+n-1) \] is the rising factorial or Pochhammer polynomial

The rising factorial satisfies the duplication formula: \[ (x)_{2n} = 4^n\left(\frac{1}{x}\right)_{n}\left(\frac{1+x}{2}\right)_{n}\] Hence \begin{align*} S= \sum_{n=0}^{\infty} \frac{(t)_{2n}}{(2)_{2n}} x^n =& \sum_{n=0}^{\infty} \frac{4^n\left(\frac{t}{2}\right)_{n}\left(\frac{1+t}{2}\right)_{n}}{4^n(1)_{n}\left(\frac{3}{2}\right)_{n}} x^n \\ =& \sum_{n=0}^{\infty} \frac{\left(\frac{t}{2}\right)_{n}\left(\frac{1+t}{2}\right)_{n}}{\left(\frac{3}{2}\right)_{n}} \frac{x^n}{n!} \quad \;\; (1)_{n} = n! \\ =& {}_{2}F_{1} \left(\frac{t}{2},\frac{t+1}{2};\frac{3}{2}, x\right) \end{align*} where ${}_{2}F_{1}$ is the Gauss hypergeometric function. ${}_{2}F_{1}$ satisfies: \[ {}_{2}F_{2} \left(a,\frac{1}{2}+a;\frac{3}{2};z^{2}\right)=\frac{1}{(2-4a)z}\left((1+z)^{1-2a}-(1-z)^{1-2a}\right)\] If we put $ a = \frac{t}{2}$ and $z^2= i^2 = -1 $ \[ \sum_{n=0}^{\infty} \frac{(t)_{2n}}{(2)_{2n}} (-1)^n = {}_{2}F_{1} \left(\frac{t}{2},\frac{t+1}{2};\frac{3}{2};-1\right) = \frac{(1+i)^{1-t}-(1-i)^{1-t}}{(2-2t)i} \tag{1} \] The derivative of the rising factorial is: \[\frac{d}{dx} (x)_{n} = (x)_{n} \left[\psi(n+x)-\psi(x)\right]\] Hence, differentiating (1) \[ \frac{d}{dt} \sum_{n=0}^{\infty} \frac{(t)_{2n}}{(2)_{2n}} (-1)^n = \frac{d}{dt} \left[ \frac{(1+i)^{1-t}-(1-i)^{1-t}}{(2-2t)i}\right] \] \begin{align*} \Longrightarrow \sum_{n=0}^{\infty} \frac{(t)_{2n}\left[\psi(2n+t)-\psi(t)\right]}{(2)_{2n}} (-1)^n =& \frac{d}{dt} \left[ \frac{(1+i)^{1-t}-(1-i)^{1-t}}{(2-2t)i}\right]\\ =& \frac{2i\left((1-i)^{1-t}-(1+i)^{1-t}\right)}{(2-2t)^2} - \frac{i\left((1-i)^{1-t}\ln(1-i)-(1+i)^{1-t}\ln(1+i)\right)}{2-2t} \end{align*} Taking the limit as $t \to 1$ \begin{align*} \sum_{n=0}^{\infty} \frac{(1)_{2n}\left[\psi(2n+1)+\gamma\right]}{(2)_{2n}} (-1)^n = &\lim_{t \to 1} \left[\underbrace{\frac{2i\left((1-i)^{1-t}-(1+i)^{1-t}\right)}{(2-2t)^2}}_{=0} - \underbrace{\frac{i\left((1-i)^{1-t}\ln(1-i)-(1+i)^{1-t}\ln(1+i)\right)}{2-2t}}_{\textrm{L'Hôpital's rule }}\right] \\ =& -\frac{\pi\ln(2)}{8} \end{align*} Using the recursion formula \[ (n+x)(x)_{n} = x(x+1)_{n}\] Hence \[ (2n+1) = \frac{(2)_{2n}}{(1)_{2n}}\] Therefore \[ \sum_{n=0}^{\infty} \frac{\left[\psi(2n+1)+\gamma\right]}{2n+1} (-1)^n = -\frac{\pi\ln(2)}{8}\] Finally, using the relation of the Harmonic number to the digamma function: \[H_x = \psi(x+1)+\gamma\] and $\displaystyle H_{0} = 0$

We can conclude \[\boxed{ \sum_{n=1}^{\infty} \frac{(-1)^nH_{2n}}{2n+1} = -\frac{\pi\ln(2)}{8}}\]