Prove the series $(-1)^k\sum_{n=0}^{\infty}{2^{n+1}(2k-1)!!\over {2n\choose n}(2n+1)(2n+3)\cdots(2n+2k+1)}=\pi-4(...)$ 
Show that for $k\ge1$,
\begin{align}
\\&\quad(-1)^k\sum_{n=0}^{\infty}{2^{n+1}(2k-1)!!\over {2n\choose n}(2n+1)(2n+3)\cdots(2n+2k+1)}\\[10pt]&=\pi-4\left(1-{1\over 3}+{1\over 5}-{1\over 7}+\cdots+{1\over 2k-1}\right).
\end{align}

I try:
The RHS sort of the favourite Leibniz $\pi$ series.
We could split into composite of fractions
${A\over 2n+1}+{B\over 2n+3}+{C\over 2n+5}+\cdots$ and then take the sum individually.
We know that 
$\sum_{n=0}^{\infty}{2^{n+1}\over {2n\choose n}(2n+1)}=\pi$
$\sum_{n=0}^{\infty}{2^{n+1}\over {2n\choose n}(2n+3)}=3\pi-8$
We don't know the general of 
$\sum_{n=0}^{\infty}{2^{n+1}\over {2n\choose n}(2n+2k+1)}=F(k)$
I am sure what to do next? Help please!
 A: Following Claude Leibovici great comment:
$$
\begin{align}
& S_k= (-1)^k\sum_{n=0}^{\infty}\frac{2^{n+1}(2k-1)!!}{{2n\choose n}(2n+1)(2n+3)\cdots(2n+2k+1)} = (-1)^k\left[\psi\left(\small\frac{2k+3}{4}\normalsize\right) - \psi\left(\small\frac{2k+1}{4}\normalsize\right)\right] \\[8mm]
& S_{k} - S_{k-1} = (-1)^k\left[\psi\left(\small\frac{2k+3}{4}\normalsize\right) - \psi\left(\small\frac{2k+1}{4}\normalsize\right)\right] - (-1)^{k-1}\left[\psi\left(\small\frac{2k+1}{4}\normalsize\right) - \psi\left(\small\frac{2k-1}{4}\normalsize\right)\right] \\[4mm]
& \qquad\qquad\space = (-1)^k\left[\psi\left(\small\frac{2k+3}{4}\normalsize\right) - \psi\left(\small\frac{2k-1}{4}\normalsize\right)\right] = (-1)^k\left[\psi\left(\small\frac{2k+3\color{red}{-4+4}}{4}\normalsize\right) - \psi\left(\small\frac{2k-1}{4}\normalsize\right)\right] \\[4mm]
& \qquad\qquad\space = (-1)^k\left[\psi\left(\small\frac{2k-1}{4}+1\normalsize\right) - \psi\left(\small\frac{2k-1}{4}\normalsize\right)\right] \qquad\qquad \small\left\{\psi(x+1)=\psi(x)+\frac{1}{x}\right\}\normalsize \\[4mm]
& \qquad\qquad\space = (-1)^k\left[\psi\left(\small\frac{2k-1}{4}\normalsize\right) + \frac{4}{2k-1} - \psi\left(\small\frac{2k-1}{4}\normalsize\right)\right] = \color{red}{4\,\frac{(-1)^k}{2k-1}} \\[8mm]
& S_k - S_0 = S_{k} -S_{k-1}+S_{k-1} \cdots -S_{1}+S_{1} -S_{0} = \sum_{n=1}^{k}\left(S_{n} -S_{n-1}\right) = 4\,\sum_{n=1}^{k}\frac{(-1)^n}{2n-1} \\[2mm]
& \small S_0=\sum_{n=0}^{\infty}{2^{n+1}\over {2n\choose n}(2n+1)}=\pi \normalsize \Rightarrow \color{red}{S_k = \pi + 4\,\sum_{n=1}^{k}\frac{(-1)^n}{2n-1}} = \pi - 4\,\left(\small 1 - \frac13 + \frac15 -\frac17 + \cdots - \frac{(-1)^k}{2k-1} \normalsize \right)
& \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}}
\end{align}
$$
A: This is probably not an answer but it is too long for a comment.
Concerning the last question $$F(k)=\sum_{n=0}^{\infty}{2^{n+1}\over {2n\choose n}(2n+2k+1)}=\frac{2 }{2 k+1}\,\,
   _3F_2\left(1,1,k+\frac{1}{2};\frac{1}{2},k+\frac{3}{2};\frac{1}{2}\right)$$ where appears the generalized hypergeometric function. Unfortunately, the expression seems to simplify only for small values of $k$
$$\left(
\begin{array}{cc}
k & F(k) \\
 0 & \pi  \\
 1 & -8+3 \pi  \\
 2 & -\frac{208}{9}+\frac{23 \pi }{3} \\
 3 & -\frac{4232}{75}+\frac{91 \pi }{5}
\end{array}
\right)$$
Concerning the general problem $$S_k=(-1)^k\sum_{n=0}^{\infty}{2^{n+1}(2k-1)!!\over {2n\choose n}(2n+1)(2n+3)\cdots(2n+2k+1)}$$ it reduces to $$S_k=(-1)^k\left(\psi \left(\frac{k}{2}+\frac{3}{4}\right)-\psi
   \left(\frac{k}{2}+\frac{1}{4}\right)\right)$$ where appears the digamma function
$$\left(
\begin{array}{cc}
 k & S_k \\
 0 & \pi  \\
 1 & \pi-4  \\
 2 & \pi-\frac{8}{3}  \\
 3 & \pi-\frac{52}{15}  \\
 4 & \pi-\frac{304}{105}  \\
 5 & \pi-\frac{1052}{315}  \\
 6 & \pi-\frac{10312}{3465}  \\
 7 & \pi-\frac{147916}{45045}  \\
 8 & \pi-\frac{135904}{45045} 
\end{array}
\right)$$
