# Four new series for $\pi$ and related identities involving harmonic numbers

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 $$H_n=\sum_{0 Namely, motivated by $$(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)?

• And to be complete, there should be something simpler than (10) featuring $(\alpha H_{2k-1}-\beta H_{k-1})$ with a similar RHS as well – obviously supposing that all these are part of a bigger (infinite?) family. Commented Oct 20, 2023 at 11:15
• Note that $K=\sqrt{3}~Gi$ where Gi is Gieseking's constant, similarly to here. Commented Oct 20, 2023 at 14:57
The "four new series for $$\pi$$" are examples of relationships between hypergeometric functions $$_pF_{p-1}$$ with rational arguments, for example, the first series is equivalent to the identity $$\, _4F_3\left(\tfrac{1}{3},\tfrac{2}{3},1,1;\tfrac{5}{4},\tfrac{3}{2},\tfrac{7}{4};\tfrac{27}{32}\right)-4 \, _4F_3\left(\tfrac{1}{3},\tfrac{2}{3},1,2;\tfrac{5}{4},\tfrac{3}{2},\tfrac{7}{4};\tfrac{27}{32}\right)+5 \, _4F_3\left(\tfrac{1}{3},\tfrac{2}{3},2,2;\tfrac{5}{4},\tfrac{3}{2},\tfrac{7}{4};\tfrac{27}{32}\right)=\frac{3\pi}{2}.$$ There is some literature on such relationships [1], but I have not (yet) found a useful connection to this case.