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The classical rational Ramanujan-type series for $1/\pi$ have the following four forms: \begin{align}\sum_{k=0}^\infty(ak+b)\frac{\binom{2k}k^3}{m^k}&=\frac{c}{\pi},\label{1}\tag{1} \\\sum_{k=0}^\infty(ak+b)\frac{\binom{2k}k^2\binom{3k}k}{m^k}&=\frac{c}{\pi},\label{2}\tag{2} \\\sum_{k=0}^\infty(ak+b)\frac{\binom{2k}k^2\binom{4k}{2k}}{m^k}&=\frac{c}{\pi},\label{3}\tag{3} \\\sum_{k=0}^\infty(ak+b)\frac{\binom{2k}k\binom{3k}k\binom{6k}{3k}}{m^k}&=\frac{c}{\pi},\label{4}\tag{4} \end{align} where $a,b,m\in\mathbb Z$, $am\not=0$ and $c^2\in\mathbb Q$. It is known that there are totally 36 such series, see, e.g., Chapter 14 of S. Cooper's book Ramanujan's Theta Functions (Springer, 2017).

For a positive integer $m$, can we find similar series for $(\log m)/\pi$? Motivated by Ramanujan-type series of the forms \eqref{1}-\eqref{4}, I have formulated the following general conjecture which involves the so-called harmonic numbers $$H_n=\sum_{0<k\le n}\frac1k\ \ \ \ (n=0,1,2,\ldots).$$

Conjecture. (i) If we have an identity \eqref{1} with $a,b,m\in\mathbb Z$, $am\not=0$ and $c^2\in\mathbb Q$, then $$\sum_{k=0}^\infty\frac{\binom{2k}k^3}{m^k}(6(ak+b)(H_{2k}-H_k)+a)=c\frac{\log|m|}{\pi}.\label{I}\tag{I}$$

(ii) If we have an identity \eqref{2} with $a,b,m\in\mathbb Z$, $am\not=0$ and $c^2\in\mathbb Q$, then $$\sum_{k=0}^\infty\frac{\binom{2k}k^2\binom{3k}k}{m^k}((ak+b)(3H_{3k}+2H_{2k}-5H_k)+a)=c\frac{\log|m|}{\pi}.\label{II}\tag{II}$$

(iii) If we have an identity \eqref{3} with $a,b,m\in\mathbb Z$, $am\not=0$ and $c^2\in\mathbb Q$, then $$\sum_{k=0}^\infty\frac{\binom{2k}k^2\binom{4k}{2k}}{m^k}(4(ak+b)(H_{4k}-H_k)+a)=c\frac{\log|m|}{\pi}.\label{III}\tag{III}$$

(iv) If we have an identity \eqref{4} with $a,b,m\in\mathbb Z$, $am\not=0$ and $c^2\in\mathbb Q$, then $$\sum_{k=0}^\infty\frac{\binom{2k}k\binom{3k}{k}\binom{6k}{3k}}{m^k}(3(ak+b)(2H_{6k}-H_{3k}-H_k)+a)=c\frac{\log|m|}{\pi}.\label{IV}\tag{IV}$$

Examples. (i) Ramanujan [Quart. J. Math. 45(1914), 350-372] found that $$\sum_{k=0}^\infty\frac{\binom{2k}k^3}{(-512)^k}=\frac{2\sqrt2}{\pi},$$ in view of this and part (i) of the above conjecture we should have $$\sum_{k=0}^\infty\frac{\binom{2k}k^3}{(-512)^k}((6k+1)(H_{2k}-H_k)+1)=\frac{2\sqrt2}{\pi}\times\frac{\log512}6=3\sqrt2\frac{\log2}{\pi}.$$

(ii) Ramanujan [Quart. J. Math. 45(1914), 350-372] found that $$\sum_{k=0}^\infty(51k+7)\frac{\binom{2k}k^2\binom{3k}k}{(-12)^{3k}}=\frac{12\sqrt3}{\pi},$$ in view of this and part (ii) of our general conjecture we should have $$\sum_{k=0}^\infty\frac{\binom{2k}k^2\binom{3k}k}{(-12)^{3k}}((51k+7)(3H_{3k}+2H_{2k}-5H_k)+51) =\frac{12\sqrt3}{\pi}\times\log 12^3=36\sqrt3\frac{\log 12}{\pi}.$$

QUESTION. Can one modify the known appraoches to classical Ramanujan-type series for $1/\pi$ to prove the above general conjecture?

It seems that our conjecture also applies to irrational Ramanujan-type series for $1/\pi$. For example, Ramanujan [Quart. J. Math. 45(1914), 350-372] found that $$\sum_{k=0}^\infty\left(k+\frac{31}{270+48\sqrt5}\right)\frac{\binom{2k}k^3}{(2^{20}/(\sqrt5-1)^8)^k}=\frac{16}{(15+21\sqrt5)\pi},$$ motivated by this and part (i) of our general conjecture we guess that \begin{align}&\sum_{k=0}^\infty\frac{\binom{2k}k^3}{(2^{20}/(\sqrt5-1)^8)^k}\left(6\left(k+\frac{31}{270+48\sqrt5}\right)(H_{2k}-H_k)+1\right) \\=&\ \frac{16}{(15+21\sqrt5)\pi}\times\log\frac{2^{20}}{(\sqrt5-1)^8}, \end{align} which can be easily checked via Mathematica.

We also have conjectures on $p$-adic congruences corresponding to identities of the forms \eqref{I}-\eqref{IV}.

Your comments are welcome!

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  • $\begingroup$ This looks great! Are you sure about the last factor 4 in (III)? All the others are "homogenous" in terms of the $H_n$, and supposing $H_0:=1$, that homogeneity makes sense. $\endgroup$
    – Wolfgang
    Dec 9, 2022 at 9:18
  • $\begingroup$ Sorry, that's a typo. It should be $H_{4k}-H_k$. I have corrected it. $\endgroup$ Dec 9, 2022 at 9:30
  • $\begingroup$ Definitely looks better. I was just noting that the exponent of $\binom{2k}k^2$ in (IV) didn't correspond to (4), I see you have also corrected that already. :) $\endgroup$
    – Wolfgang
    Dec 9, 2022 at 9:32
  • $\begingroup$ I also conjecture that $$\sum_{k=1}^\infty\frac{81^k((35k-8)(H_{4k-1}-H_{k-1})-35/4)}{k^3\binom{2k}k^2\binom{4k}{2k}}=12\pi^2\log3+39\zeta(3).$$ For more such series, one may consult my recent preprint available from arxiv.org/abs/2210.07238. $\endgroup$ Dec 9, 2022 at 10:35

1 Answer 1

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"Can one modify the known appraoches to classical Ramanujan-type series for $1/\pi$ to prove the above general conjecture?" -- Yes.

Actually much more can be proved.


We use classical notations (see for example, Pi and AGM by Browein, Chap. 5.5). For $0\leq s < 1/2$, let $$K_s(k) = \frac{\pi}{2} {_2F_1}(\frac{1}{2}-s,\frac{1}{2}+s;1;k^2) \qquad E_s(k) = \frac{\pi}{2} {_2F_1}(-\frac{1}{2}-s,\frac{1}{2}+s;1;k^2)$$ $k' = \sqrt{1-k^2}$, $K_s'(k) = K_s(k')$. $$\alpha_s(r) = \frac{\pi}{4K_s^2} \frac{\cos \pi s}{1+2s} - \sqrt{r}(\frac{E_s}{K_s}-1)$$ for positive rational $N$, let $k_{N}$ be such that $K_s'(k_{N}) = \sqrt{N}K_s(k_{N})$. It can be shown $$\tag{1}\frac{\cos \pi s}{\pi (1+2s)} = \frac{\sqrt{N}k_N k_N'^2}{1+2s} \frac{4K_s}{\pi^2} \frac{d K_s}{dk} + [\alpha_s(N)-\sqrt{N}k_N^2] \frac{4K_s^2}{\pi^2}$$ $$\tag{2}0 = \frac{\sqrt{N}k_N k_N'^2}{2(1+2s)} \frac{d}{dk} (K_s K'_s) + [\alpha_s(N)-\sqrt{N}k_N^2] K_s K'_s$$

When $s\in \{0,1/3,1/6,1/4\}$, $N$ positive rational, all quantities in $(1), (2)$ except $K, K'$, are algebraic number: because they are values of modular (or derivatives thereof) functions at CM points. These are of course, well-known results.


For the rest, assuming $0\leq s < 1/2$ is enough. If $0\leq k < \frac{1}{\sqrt{2}}$, we have

$$\tag{*}\begin{align*}K_s(k)^2 &= \frac{\pi^2}{4}\sum_{n\geq 0} c_n (2kk')^{2n} \\ K_s(k) K'_s(k) &= \frac{\pi \cos s\pi}{4}\sum_{n\geq 0} c_n (d_n - 2\log(2kk')) (2kk')^{2n} \\ K'_s(k)^2 + K_s(k)^2&= \frac{\cos^2(\pi s)}{4} \sum_{n\geq 0} c_n \left(-4 d_n \log (2kk')+e_n+4\log^2(2kk')\right) (2kk')^{2n} \end{align*}$$ here, in terms of Pochhammer symbol and polygamma function, $$\begin{aligned} c_n &= \frac{(1/2-s)_n (1/2+s)_n (1/2)_n}{(n!)^3} \\ d_n &= -\psi(n-s+\frac{1}{2})-\psi(n+s+\frac{1}{2})+3 \psi(n+1)-\psi(n+\frac{1}{2})\\ e_n &= \text{ certain expression in digamma and trigamma}\end{aligned}$$ An intuitive explanation of all formulas $(*)$: LHS are solutions of a 3rd order ODE, whose indicial equation at origin has triple root. The $\log, \log^2$ comes from general theory of Frobenius method, and polygamma are typical in the expansion of such solutions.

Inserting the first series into $(1)$, gives our familiar $1/\pi$-formula: $$\frac{\cos \pi s}{\pi (1+2s)} = \sum_{n\geq 0} c_n (A_{s,N} n + B_{s,N}) (2kk')^{2n}$$ with $A_{s,N} = \frac{\sqrt{N}}{1+2s} (-k^2+k'^2)$ and $B_s(N) = \alpha_s(N) - \sqrt{N}k_N^2$.

If one uses instead the expansion of $K_s K'_s$, and plug it into $(2)$: $$ 0 = -2\log(2kk') \frac{\cos \pi s}{\pi(1+2s)} + \sum_{n\geq 0} c_n (2kk')^{2n} (-A_{s,N} + (n A_{s,N} + B_{s,N}) d_n)$$

When $s=0$, this reduces to conjecture (I) of OP: $d_n =3\log 4 -6 \left(H_{2 n}-H_n\right)$, so above expression becomes $$\frac{1}{\pi} \log [4^3(2kk')^2] = \sum_{n\geq 0} c_n (2kk')^{2n} [A_{0,N} + 6(H_{2n}-H_n) (nA_{0,N} + B_{0,N})]$$ matching the form conjectured by OP.

Conjectures (II), (III), (IV) follow from $s=1/6, 1/4, 1/3$ respectively. Q.E.D.

If one uses the expansion for $K'_s(k)^2 + K_s(k)^2$, one obtains similar series involving second order harmonic numbers.

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  • $\begingroup$ Very nice derivation! +1 $\endgroup$ Dec 21, 2022 at 5:03

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