Recently, I discovered the following four new (conjectural) series for $\pi$: \begin{align}\sum_{k=1}^\infty\frac{(5k^2-4k+1)8^k\binom{3k}k}{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}}&=\frac{3\pi}2,\tag{1} \\[8pt]\sum_{k=1}^\infty\frac{(35k^2-29k+6)3^k\binom{3k}k}{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}}&=\sqrt3\,\pi,\tag{2} \\[8pt]\sum_{k=1}^\infty\frac{(40k^2-20k+3)2^k\binom{4k}{2k}} {k(4k-1)(4k-3)\binom{3k}k\binom{6k}{3k}}&=\frac{\pi}2,\tag{3} \\[8pt]\sum_{k=1}^\infty\frac{(64k^2-48k+7)3^k\binom{4k}{2k}} {k(4k-1)(4k-3)\binom{3k}k\binom{6k}{3k}}&=\frac{4\pi}{3\sqrt3}.\tag{4} \end{align} The series in $(1)$–$(4)$ have converging rates $27/32$, $81/256$, $2/27$ and $1/9$ respectively, and so it is easy to check the identities $(1)$–$(4)$ numerically.
I also consider some variants of $(1)$–$(4)$ involving the harmonic numbers $$H_n=\sum_{0<k\le n}\frac1k\ \ (n=0,1,2,\ldots).$$ Namely, motivated by $(1)$-$(4)$, I conjecture the following identities: \begin{align}\sum_{k=1}^\infty\frac{\binom{3k}k8^k((5k^2-4k+1)(2H_{2k-1}-3H_{k-1})-8k+2)}{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}} &=3\pi\log2-12G,\tag{5} \\\sum_{k=1}^\infty\frac{\binom{3k}k8^k((5k^2-4k+1)(6H_{4k-1}-7H_{2k-1})+22k-10)}{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}} &=3\pi\log2+24G,\tag{6} \end{align} \begin{align} &\sum_{k=1}^\infty\frac{\binom{3k}k8^k\left((5k^2-4k+1)(2H_{3k-1}-2H_{2k-1}-H_{k-1})+\frac{2(219k^3-249k^2+87k-10)}{3k(3k-1)}\right)}{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}} \\&\qquad\qquad=\frac{9}4\pi^2,\tag{7}\end{align} \begin{align} &\sum_{k=1}^\infty\frac{\binom{3k}k3^k\left((35k^2-29k+6)(H_{2k-1}-H_{k-1})-\frac{4(4k-1)(7k-3)}{5(2k-1)}\right)} {k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}} \\&\qquad=\frac{3}{10}(2\pi\sqrt3\log3-9K),\tag{8} \end{align} \begin{align}&\sum_{k=1}^\infty\frac{\binom{3k}k3^k\left((35k^2-29k+6)(H_{4k-1}-H_{2k-1})+\frac{2(42k^2-36k+7)}{5(2k-1)}\right)} {k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}} \\&\qquad=\frac{99}{10}K-\frac{\pi}5\sqrt3\log3,\tag{9} \end{align} \begin{align} &\sum_{k=1}^\infty\frac{\binom{4k}{2k}2^k((40k^2-20k+3)(2H_{6k-1}-H_{3k-1}-H_{k-1})-32k+4)} {k(4k-1)(4k-3)\binom{3k}k\binom{6k}{3k}} \\&\qquad=2G+\frac{\pi}2\log2,\tag{10} \end{align} \begin{align} &\sum_{k=1}^\infty\frac{\binom{4k}{2k}3^k\left((64k^2-48k+7)(5H_{2k-1}-4H_{k-1})-\frac{2(6k-1)(24k-13)}{2k-1}\right)} {k(4k-1)(4k-3)\binom{3k}k\binom{6k}{3k}} \\&\qquad=\frac43\pi\sqrt3\log3-9K,\tag{11} \end{align} and \begin{align} &\sum_{k=1}^\infty\frac{\binom{4k}{2k}3^k\big((64k^2-48k+7)(10H_{6k-1}-5H_{3k-1}-3H_{k-1})-\frac{8(96k^2-78k+17)}{6k-3}\big)} {k(4k-1)(4k-3)\binom{3k}k\binom{6k}{3k}} \\&\qquad=\frac49\pi\sqrt3\log3+32K,\tag{12} \end{align} where $$G:=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2} \ \ \text{and}\ \ \ K:=\sum_{k=1}^\infty\frac{(\frac k3)}{k^2}=\sum_{n=0}^\infty\left(\frac1{(3n+1)^2}-\frac1{(3n+2)^2}\right).$$
QUESTION. Can one prove the identities $(1)$-$(12)$ via known tools (including the WZ pairs and hypergeometric transformation formulas)?
Your commments are welcome!
