Are these fast convergent series for $\log(2)$, $\log(3)$ and $\log(5)$ already known and proven?

Question

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Now that some of the previously MSE formulae that I left here have been applied Dec.2023 to compute high precision record values ($10^{12}$ decimal digits) of trascendental constants $\Gamma(1/3)$ (Eq.B.IV.8) and $\Gamma(1/4)$ (Eq.A.IIIa.4 and Eq.A.III.3) as it is reported in (a) and (b)-(c), I have undertaken the search for highly efficient series $s$ to calculate $\log(2)$, $\log(3)$ and $\log(5)$ that are computed by the binary splitting method. I have found one conjectured series for $s=\log(3)$, two for $s=\log(2)$ and one for $s=\log(5)$. I am not sure if any of them has been already published, so the question is very simple and the same as the former note: Is any of these series known?. If they are not, I am interested to know a proof for them. Perhaps by means of Wilf-Zeilberger pairs. (See answer below, updated on Nov 13 2024). Any suggestions on this way are welcome.

We use the following notation, where the constant $s$ is expressed as $$s=\sum_{n=1}^\infty\,\rho^n\cdot\frac{p(n)}{r(n)}\cdot\left[\begin{matrix} a & b & c & ... & z \\ A & B & C & ... & Z \\ \end{matrix}\right]_n=\sum_{n=1}^\infty\frac{p(n)}{r(n)}\cdot\prod_{k=1}^n\frac{r(k)}{q(k)}$$ here $p(n),q(n),r(n)$ are polynomials non vanishing for $n\in\mathbb{N}$, $q(n)$ and $r(n)$ have the same degree $d$ and the convergence ratio $|\rho|$ is the absolute value of the ratio of the leading terms of $r(n)$ and $q(n)$. The ratio of products of Pochhammer's symbols (rising factorials) is written as $$\left[\begin{matrix} a & b & c & ... & z \\ A & B & C & ... & Z \\ \end{matrix}\right]_n=\frac{(a)_n(b)_n(c)_n ... (z)_n}{(A)_n(B)_n(C)_n ... (Z)_n} $$ where the degree $d$ is the number of elements in a row (they are the same for both rows) and $$(w)_n = \frac{\Gamma(w+n)}{\Gamma(w)}=w(w+1)(w+2)...(w+n-1)$$

The computational speed is measured through the binary splitting cost $$ C_s = - \frac{4d}{\log|\rho|}.$$ This allows to (asymptotically) rank, classify and compare different hypergeometric-type algorithms by performance.

I came across these expressions. Three of them look pretty simple Ramanujan type formulae,

A. For $\log(3)$

$$\begin{equation*}\log(3)=\sum_{n=1}^\infty\left(\frac{1}{3^{5}}\right)^n\cdot\frac{88\,n-14}{n(2n-1)}\cdot\left[\begin{matrix} 1&\frac{1}{2}\\ \frac{1}{6}&\frac{5}{6}\\ \end{matrix}\right]_n\tag{1}\label{1} \end{equation*}$$ It has a binary splitting cost $C_s=\frac{8}{5\,\log(3)}=1.4638..$. In preliminary tests this expression performs faster than the fastest known series for such constant that is based on a 4-term Machin-like arcotanh formula with arguments 251, 449, 4801 and 8749. See Table 1 here.

B. For $\log(2)$

$$\begin{equation*}\small{\log(2)}=\small{\sum_{n=1}^\infty\left(\frac{1}{3^{3}\cdot2^{13}}\right)^n\cdot\frac{P(n)}{3n(2n-1)(3n-1)(3n-2)}\cdot\left[\begin{matrix} 1&\frac{1}{2}&\frac{1}{3}&\frac{2}{3}\\ \frac{1}{12}&\frac{5}{12}&\frac{7}{12}&\frac{11}{12}\\ \end{matrix}\right]_n}\tag{2}\label{2} \end{equation*}$$ where $$\small{P(n)}=\small{686430\,n^3 - 742257\,n^2 + 223397\,n - 13858}$$

It has a binary splitting cost $C_s=\frac{16}{\log(3^{3}\cdot2^{13})}=1.3001..$. Preliminary tests show that this series performs slightly faster than the fastest known series for such constant that is based on a 3-term Machin-like arcotanh formula with arguments 26, 4801 and 8749. See here.

The third one is,

$$\begin{equation*}\log(2)=\frac{1}{2}\,\sum_{n=1}^\infty\left(\frac{1}{3^{5}\cdot2^{4}}\right)^n\cdot\frac{1794\,n-297}{n(2n-1)}\cdot\left[\begin{matrix} 1&\frac{1}{2}\\ \frac{1}{6}&\frac{5}{6}\\ \end{matrix}\right]_n\tag{3}\label{3} \end{equation*}$$ This expression has a binary splitting cost $C_s=\frac{8}{\log(3^{5}\cdot2^{4})}=0.96786..$. Preliminary tests show that this series performs much faster than the fastest known series for such constant that is based on the mentioned 3-term Machin-like formula.

$\log(2)$ is an important fundamental constant and this last expression, being very efficient, should be taken as a standard high precision formula for this constant to be included inside mathematical software whenever it is implemented as a binary splitting algorithm. In fact this formula will be part of FLINT as it is reported here

C. For $\log(5)$

$$\begin{equation*}\log(5)=\sum_{n=1}^\infty\left(\frac{-1}{3^{3}\cdot5^2}\right)^n\cdot\frac{364\,n-62}{-n(2n-1)}\cdot\left[\begin{matrix} 1&\frac{1}{2}\\ \frac{1}{6}&\frac{5}{6}\\ \end{matrix}\right]_n\tag{4}\label{4} \end{equation*}$$ It has a binary splitting cost $C_s=\frac{8}{\log(675)}=1.2280..$. Preliminary tests show that this series performs pretty faster than the fastest known series for such constant that is based on a 4-term Machin-like formula (a linear combination of arcotanhs with arguments 251, 449, 4801 and 8749).

Eqs.(3-4) allow also to get high precision values of $\log(10)$ that is an important constant in numerical analysis.

Q: Is any of Eqs.(1-4) known? and if they are not, would it be possible to get the proofs?

I would like to thank Dr. J. Guillera for encouraging me to investigate finding these formulas. To H. Cohen and the PARI GP team at the University of Bordeaux for this excellent parallelizable search tool and to Jordan Ranous at Storage Review for providing me with high-performance multicore facilities.

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Updated on Feb.16.2024

Fortunately for three of these series, Eqs. (1), (3) and (4), I have found a proof using classical methods which I placed in the Answers section below. (I doubt that there is a proof based on modular forms, this transformation of coefficients for the logarithm is not in PSL2(ℝ).)

Eq.(3) (now proven) was used on Feb.12 to double up to $3$ x $10^{12}$ the known number of decimal digits places of log(2), Eq.(2) (unproven) was used for verifying, as it is reported here. Details are found in this log. Formulas below are unpublished and are just placed here at MO.

Updated on Jul.19.2024

At the end of my response in the answer's section below, I have placed a new proof based on some closed forms found in a recently published work (2024) by by Zhi-Wei Sun and Yajun Zhou arXiv:2401.14197v1 [math.CA], 25 Jan 2024.

Updated on Nov.13 and Nov.21.2024

D. New series for $\log(2)$ and $\log(3)$

The following 3 additional series for $\log(2)$ are even more efficient than Eq.(2) under the application of the binary splitting algorithm,

$$\begin{equation*}\small{\log(2)}=\small{\sum_{n=1}^\infty\left(\frac{1}{2^4\cdot3^2\cdot5^5}\right)^n\cdot\frac{P(n)}{4n(2n-1)(4n-1)(4n-3)}\cdot\left[\begin{matrix} 1&\frac{1}{2}&\frac{1}{4}&\frac{3}{4}\\ \frac{1}{10}&\frac{3}{10}&\frac{7}{10}&\frac{9}{10}\\ \end{matrix}\right]_n}\tag{5}\label{5} \end{equation*}$$ where $$\small{P(n)} = \small{3927264\,n^3 - 4300512\,n^2 + 1209726\,n - 81891}$$ It has a binary splitting cost $Cs=1.2291..$ performing very fast. (This series was privately communicated by MO user @xiaoshuchong).

The next series was found using the PSLQ algorithm. It has a cost $Cs=1.1335..$ providing the currently 2nd fastest formula to compute this constant. This is a good candidate to be implemented in y-cruncher software as a digits verification algorithm for $\log(2)$

$$\begin{equation*}\small{\log(2)}=\small{\sum_{n=1}^\infty\left(\frac{1}{2^4\cdot3^3\cdot5^5}\right)^n\cdot\frac{P(n)}{2n(2n-1)(6n-1)(6n-5)}\cdot\left[\begin{matrix} 1&\frac{1}{2}&\frac{1}{6}&\frac{5}{6}\\ \frac{1}{10}&\frac{3}{10}&\frac{7}{10}&\frac{9}{10}\\ \end{matrix}\right]_n}\tag{6}\label{6} \end{equation*}$$ where $$\small{P(n)}=\small{13885704\,n^3 - 15397068\,n^2 + 4353342\,n - 295245}$$

Also, this one was found (and proven) using WZ pairs

$$\begin{equation*}\small{\log(2)}=\small{\sum_{n=1}^\infty\left(\frac{1}{2^4\cdot3^3\cdot7^7}\right)^n\cdot\frac{P(n)}{Q(n)}\cdot\left[\begin{matrix} 1&\frac12&\frac14&\frac34&\frac16&\frac56\\ \frac1{14}&\frac3{14}&\frac5{14}&\frac9{14}&\frac{11}{14}&\frac{13}{14}\\ \end{matrix}\right]_n}\tag{7}\label{7} \end{equation*}$$ where $$\small{P(n)}=\small{81969540480\,n^5 - 169950180480\,n^4 + 126495134424\,n^3 }$$$$\small{- 40884797604\,n^2 + 5510613042\,n - 226846575}$$$$\small{Q(n)}=\small{4n(2n-1)(4n-1)(4n-3)(6n-1)(6n-5)}$$

It has a binary splitting cost $C_s=1.21890$

Finally, for $\log(3)$, the following series was found with PSLQ. It has a cost $C_s=1.64589...$. This series performs slightly better than Machin-like arcotanh formulas being the 2nd fastest series currently known. $$\begin{equation*}\small{\log(3)}=\small{\sum_{n=1}^\infty\left(\frac{3}{2^4\cdot5^5}\right)^n\cdot\frac{P(n)}{n(2n-1)(6n-1)(6n-5)}\cdot\left[\begin{matrix} 1&\frac{1}{2}&\frac{1}{6}&\frac{5}{6}\\ \frac{1}{10}&\frac{3}{10}&\frac{7}{10}&\frac{9}{10}\\ \end{matrix}\right]_n}\tag{8}\label{8} \end{equation*}$$ where $$\small{P(n)}=\small{141168\,n^3 - 158016\,n^2 + 44804\,n - 3040}$$

It is interesting that these known fastest $\log(2)$ series, Eqs.(3), (5), (6) and (7), and $\log(3)$ series, Eqs. (1) and (8), are all proven by means of Wilf-Zeilberger $(F,G)$ pairs starting from the same WZ seed and base companion $F(n,k)$ by using a method indicated in this MO question. The proofs are placed as an update to my answer below.

Answer 1

A few superficial low-tech remarks that probably aren't useful at all, just in case they spark more useful ideas for others.

1. It's easy to rewrite the "quotient of product of rising factorials" factor that appears in A, B2, C in terms of factorials and powers of (in this case) 2 and 3, and doing so results in smaller powers of 2 and 3 appearing overall. In particular, this factor equals $2^{2n}3^{3n}(2n)!^2(3n)!/n!(6n)!$. I suspect that for most purposes the Pochhammer expression is better, but it is nice to cancel out some of those 2s and 3s. (Of course you could do the same for B1 as well. I haven't done the calculations. I suspect it results in fewer 2s and 3s there too.)

2. Those $An-B$ factors are suspiciously close to multiples of $6n-1$. It feels natural to rewrite them as $p(6n-1)+q(2n-1)$; the coefficients come out quite nice.

3. When we write things this way, the value it's natural to think of as "the value of $x$ in the Taylor series" is in each case $\pm r^2$ where $r$ is a simple rational number. We have a minus sign in the $\log 5$ case only. Could there maybe be a formula for $\log p$ that needs $x$ to involve a $(p-1)$th root of unity?

More concretely, if we write

$$\begin{align} a(x) & = \sum_{n=1}^{\infty}\frac{(2n-2)!(2n-1)!(3n)!}{n!(6n-2)!}x^{2n} \\ b(x) & = \sum_{n=1}^{\infty}\frac{(2n-1)!(2n)!(3n)!}{n!(6n)!}x^{2n} \end{align}$$

then Jorge's formulae can be written as

$$\begin{align} \log 2 & = 100 a(1/6\cdot 1) - 3 b(1/6\cdot 1) \\ \log 3 & = 10 a(2/3\cdot -1) - 2 b(2/3\cdot -1) \\ \log 5 & = -40 a(2/5\cdot i) - 4 b(2/5\cdot i). \end{align}$$

In the unlikely event that this roots-of-unity stuff isn't nonsense we might actually prefer to write that last one as

$$\log 5 = -20 a(2/5\cdot i) - 20 a(2/5\cdot -i) - 2 b(2/5\cdot i) - 2 b(2/5\cdot -i).$$

But, I assume to no one's surprise, it doesn't in fact seem to be the case (according to my simple-minded experiments, which I would not advise anyone to trust very much) that

$$\log 7 = p [a(r\omega)+a(r\bar\omega)] + q [b(r\omega)+b(r\bar\omega)]$$

for small integer $p,q$, small rational $r$, and $\omega,\bar\omega$ the two primitive (7-1)th roots of unity.

[EDITED to add:] A further similarly-superficial remark about those factorial-y formulae for $a,b$: we can rewrite them in various ways by adding/cancelling factors "at the ends of the factorials". For instance, in the expression for $b(x)$ we can turn $2n,3n,n,6n$ into $2n-1,3n-1,n-1,6n-1$ which arguably gains a little bit of symmetry. We can likewise make lots of tweaks to $a(x)$ but I haven't spotted any that make the formula obviously simpler or more elegant. It doesn't seem likely that any of this sort of noodling leads anywhere important.

As for the B1 formula, if my calculations are correct (which they might be, I guess) it's equivalent to

$$\log 2=\frac13\sum2^{-3n}\frac{686430n^3-742257n^2+223397n-13858}{n(2n-1)(3n-1)(3n-2)}\frac{(3n)!(4n)!(6n)!}{n!(12n)!}$$ (so indeed those factors of 2 and 3 get simpler when we write things this way) where I suspect the $n$ in the denominator in the middle is really either a $2n$ or a $3n$ (with compensating factor in the numerator) or maybe even both. Or maybe we should think of the denominator as $6n(6n-1)(6n-3)(6n-4)/72$. I hoped, by analogy with the other formulae, that the numerator might turn out to be some smallish-coefficient combination of things like $(6n-1)(6n-2)(6n-3)$ but haven't actually found such a combination and am not sure whether e.g. we should expect something like $(12n-1)(12n-2)(12n-3)$ in there. If we write the numerator as a linear combination of $(kn-1)(kn-2)(kn-3)$ for $k=12,6,3,2$ then the resulting coefficients are not very encouraging.

Answer 2

UPDATE 07.2024

At the bottom of this answer I have placed a new proof based on a recently published work (2024). It applies some closed forms derived by Zhi-Wei Sun and Yajun Zhou arXiv:2401.14197v1 [math.CA], 25 Jan 2024.

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By naming $S_2,S_1$ and $S_3$ the rhs series in Eqs.(1), (3) and (4) respectively, a proof that $S_\ell=\log\left(\frac{7+\ell}{5-\ell}\right)$ for $\ell=1,2,3$ is built transforming them to an integral form of this type $$\begin{equation*}S_\ell=\int_0^1\frac{P(x)}{Q(x)}\frac{dx}{\sqrt{1-x}}\tag{I}\label{I} \end{equation*}$$ which must be solved with $P(x)$ and $Q(x)$ polynomials to be found.

We need first to write these series starting from $n=0\,$ as $$\begin{equation*}S_\ell=\frac{1}{c}\cdot\sum_{n=0}^\infty\,\rho^n\cdot\frac{a\,n+b}{(6n+1)(6n+5)}\cdot\left[\begin{matrix} 1&\frac{1}{2}\\ \frac{1}{6}&\frac{5}{6}\\ \end{matrix}\right]_n\tag{II}\label{II} \end{equation*}$$ where $$\left[\begin{matrix} \ell & a & b & c & \rho \\ 1 & 1794 & 1497 & 432 & 2^{-4}\cdot3^{-5} \\ 2 & 176 & 148 & 27 & 3^{-5} \\ 3 & 728 & 604 & 75 & -3^{-3}\cdot5^{-2} \end{matrix}\right]$$ Using the duplication formula for $\Gamma$ function, the hypergeometric motive is written in this form $$\left[\begin{matrix} 1&\frac{1}{2}\\ \frac{1}{6}&\frac{5}{6}\\ \end{matrix}\right]_n = 108^n\cdot\frac{(2n)!^2\cdot(3n)!}{n!\cdot(6n)!}=$$ $$=\frac{108^n}{16^n}\cdot\binom{3n-1/2}{2n}^{-1}=\left(\frac{27}{4}\right)^n\cdot\frac{\Gamma(2n+1)\,\Gamma(n+1/2)}{\Gamma(3n+1/2)}$$ So Eq.(II) can be expressed as $$\begin{equation*}S_\ell=\sum_{n=0}^\infty\,\left(\frac{27}{4}\cdot\rho\right)^n\cdot\,G(n)\cdot\frac{\Gamma(2n+1)\,\Gamma(n+1/2)}{\Gamma(3n+1/2)}\tag{III}\label{III}\end{equation*}$$ where $$G(n)=\frac{1}{c}\cdot\frac{a\,n+b}{(6n+1)(6n+5)}$$ This expression is written as a special partial fraction $$G(n)=A_1\cdot\frac{1}{3n+1/2}+A_2\cdot\frac{2n+1}{(3n+1/2)(3n+3/2)}+A_3\cdot\frac{(2n+1)(2n+2)}{(3n+1/2)(3n+3/2)(3n+5/2)}$$ giving the following values $$\left[\begin{matrix} \ell & A_1 & A_2 & A_3 \\ 1 & \frac{25}{72} & -\frac{1}{192} & \frac{1}{192}\\ 2 & \frac{5}{9} & -\frac{1}{18}& \frac{1}{18}\\ 3 & \frac{4}{5} & \frac{1}{25} & -\frac{1}{25} \end{matrix}\right]$$ We now express the series by naming $$B_{n,k}=B(n+\frac{1}{2},2n+k)=\frac{\Gamma(n+\frac{1}{2})\,\Gamma(2n+k)}{\Gamma(3n+k+\frac{1}{2})}=\int_0^1x^{2n+k-1}(1-x)^{n-\frac{1}{2}}dx$$ for $k=1,2,3$ this gives $$S_\ell=\sum_{n=0}^\infty\,\left(\frac{27}{4}\cdot\rho\right)^n\left[A_1\cdot B_{n,1}+A_2\cdot B_{n,2}+A_3\cdot B_{n,3}\right]$$ By dominated convergence sum and integral commute $$S_\ell=\int_0^1(A_1+A_2\,x+A_3\,x^2)\cdot\sum_{n=0}^\infty\,\left(\frac{27}{4}\cdot\rho\cdot\,x^2(1-x)\right)^n \frac{dx}{\sqrt{1-x}}$$ Finally by replacing $A_1, A_2, A_3$ and $\rho$ in $$S_\ell=\int_0^1\frac{A_1+A_2\,x+A_3\,x^2}{1-\frac{27}{4}\rho\cdot\,x^2(1-x)}\cdot\frac{dx}{\sqrt{1-x}}$$ we get Eq.(I) with the following polynomials$$\left[\begin{matrix} \ell & P(x) & Q(x) \\ 1 & 3x^2-3x+200 & (x+8)\cdot(x^2-9x+72)\\ 2 & 2x^2-2x+20& (x+3)\cdot(x^2-4x+12)\\ 3 & 4x+16 & x^2+4x+20 \end{matrix}\right]$$ I guess there are some variable transformations like $x=1-\phi(y)^2$ to put these integrals in simpler form to be evaluated, but I left this step to the interested reader. We use Mathematica to solve these expressions symbolically. We got this output,

Finally $S_\ell=\log\left(\frac{7+\ell}{5-\ell}\right)$ for $\ell=1,2,3$ and these fast series are proven.

NOTE: This method can be also used to prove many relationships (including other constants beyond logarithms) based on hypergeometric-type series whenever the hypergeometric motive can be put in the following form $$\left[\begin{matrix} a & b & c & ... & z \\ A & B & C & ... & Z \\ \end{matrix}\right]_n=4^{\nu n}\cdot\alpha^n\cdot\frac{(\nu\,n)!(2m\,n)!(N\,n)!}{(m\,n)!(2Nn)!}$$ $$=\alpha^n\cdot\frac{\Gamma(\nu n+1)\cdot\Gamma(m n+1/2)}{\Gamma(N n+1/2)}$$ for integers $m\ge0, \nu>0$ and $N = m + \nu$, with $\alpha=\frac{N^N}{m^m\nu^{\nu}}$ and convention $0^0=1$. In this case we get the series converted into $$\int_0^1\frac{P(x)}{1-\alpha\cdot\rho\cdot x^\nu\cdot(1-x)^m}\cdot\frac{dx}{\sqrt{1-x}}$$ for some polynomial $P(x)$ of degree less than $N$.

=============================================================== UPDATE 07.2024. NEW PROOFS

The logarithm series $$\begin{equation*}\log\,p=\frac{1}{\gamma}\cdot\sum_{n=1}^\infty\,\rho^n\cdot\frac{\alpha\,n+\beta}{n\,(2n-1)}\cdot\left[\begin{matrix} 1&\frac{1}{2}\\ \frac{1}{6}&\frac{5}{6}\\ \end{matrix}\right]_n \end{equation*}$$ $$\begin{equation*}\log\,p=\frac{1}{\gamma'}\cdot\sum_{n=0}^\infty\,\rho^n\cdot\frac{\alpha'n+\beta'}{(6n+1)(6n+5)}\cdot\left[\begin{matrix} 1&\frac{1}{2}\\ \frac{1}{6}&\frac{5}{6}\\ \end{matrix}\right]_n \end{equation*}$$can be proven for these sets of parameters $$\left[\begin{matrix} p & \alpha & \beta & \gamma & \alpha' & \beta' & \gamma' & \rho & x \\ 2 & 1794 & -297 & 598 & 499 & 144 & 2 & 2^{-4}\cdot3^{-5} & 3 \\ 3 & 88 & -14 & 1 & 176 & 148 & 27 & 3^{-5} & 2 \\ 5 & -364 & 62 & 1 & 728 & 604 & 75 & -3^{-3}\cdot5^{-2} & 2i \\ 7 & 312 & -16 & 81 & 468 & 444 & 49 & 3^3\cdot2^{-2}\cdot7^{-2} & \frac{4}{3} \\ 10 & -126 & 23 & 2 & 1134 & 927 & 80 & -2^{-4}\cdot5^{-1} & i\frac{\sqrt{15}}{3} \end{matrix}\right]$$ where $p=2,3,5$ series perform faster than Machin-type formulas but $p=7,10$ do not. (Anyway $p=10$ series is more efficient than adding the fast series $p=2$ and $p=5$).

From Table 3 in Zhi-Wei Sun and Yajun Zhou arXiv:2401.14197v1 article we take the rows A.1, B.1, C.1 and D.1 and we write them as $$\Phi_A(x)=\sum_{n=1}^\infty\frac{\binom{2n}{n}}{n\,\binom{3n}{n}\binom{6n}{3n}}\cdot\left[\frac{4}{x(1-x^2)}\right]^{2n}=\sum_{n=1}^\infty\frac{\,\rho^n}{n}\cdot\left[\begin{matrix} 1&\frac{1}{2}\\ \frac{1}{6}&\frac{5}{6}\\ \end{matrix}\right]_n$$ $$\Phi_B(x)=\sum_{n=0}^\infty\frac{\binom{2n}{n}}{(6n+1)\binom{3n}{n}\binom{6n}{3n}}\cdot\left[\frac{4}{x(1-x^2)}\right]^{2n}=\sum_{n=0}^\infty\,\frac{\rho^n}{6n+1}\cdot\left[\begin{matrix} 1&\frac{1}{2}\\ \frac{1}{6}&\frac{5}{6}\\ \end{matrix}\right]_n$$ $$\Phi_C(x)=\sum_{n=0}^\infty\frac{\binom{2n}{n}}{(6n+5)\binom{3n}{n}\binom{6n}{3n}}\cdot\left[\frac{4}{x(1-x^2)}\right]^{2n}=\sum_{n=0}^\infty\,\frac{\rho^n}{6n+5}\cdot\left[\begin{matrix} 1&\frac{1}{2}\\ \frac{1}{6}&\frac{5}{6}\\ \end{matrix}\right]_n$$ $$\Phi_D(x)=\sum_{n=1}^\infty\frac{\binom{2n}{n}}{(2n-1)\binom{3n}{n}\binom{6n}{3n}}\cdot\left[\frac{4}{x(1-x^2)}\right]^{2n}=\sum_{n=1}^\infty\,\frac{\rho^n}{2n-1}\cdot\left[\begin{matrix} 1&\frac{1}{2}\\ \frac{1}{6}&\frac{5}{6}\\ \end{matrix}\right]_n$$ with $$\Phi_A(x)=\frac{1}{1-3x^2}\left[6x\tanh^{-1}\left(\small{\frac{1}{x}}\right)+(3x^2-2)\cdot\Phi_L(x)\right]$$ $$\Phi_B(x)=\frac{3x(1-x^2)}{2(1-3x^2)}\left[\tanh^{-1}\left(\small{\frac{1}{x}}\right)-\frac{x}{2}\cdot\Phi_L(x)\right]$$ $$\Phi_C(x)=\frac{3x(1-x^2)}{2(3x^2-1)}\left[(9x^4-9x^2+1)\tanh^{-1}\left(\small{\frac{1}{x}}\right)+\frac{x\,(9x^4-15x^2+5)}{2}\cdot\Phi_L(x)\right]$$ $$\Phi_D(x)=\frac{1}{2(1-x^2)(1-3x^2)}\left[\frac{3x^4-3x^2+2}{x}\tanh^{-1}\left(\small{\frac{1}{x}}\right)+\frac{x^2\,(3x^2-5)}{2}\cdot\Phi_L(x)\right]$$ where $$\Phi_L(x)=i\cdot\frac{1}{\sqrt{3x^2-4}}\log\left(\frac{x^2-2+i\,\sqrt{3x^2-4}}{x^2-2-i\,\sqrt{3x^2-4}}\right)$$

With simple algebra we can build the $\log\,p$ series above by taking $$\rho=\frac{4}{27}\cdot\frac{1}{x^2(1-x^2)^2}$$ and making these linear combinations of the A.1 and D.1 series or the B.1 and C.1 series, which gives $$\log\,p=-\frac{\beta}{\gamma}\cdot\Phi_A(x)+\frac{\alpha+2\beta}{\gamma}\cdot\Phi_D(x)$$

$$\log\,p=\frac{6\beta'-\alpha'}{24\gamma'}\cdot\Phi_B(x)+\frac{5\alpha'-6\beta'}{24\gamma'}\cdot\Phi_C(x)$$

For $\rho>0$, $x$ is obtained as the real root of $x(1-x^2)=\frac{2}{3}\sqrt{\frac{1}{3\rho}}$. For $\rho<0$, $x=i\,y$ where $y$ is got as the real root of $y(1+y^2)=\frac{2}{3}\sqrt{\frac{1}{3|\rho|}}$. Last column in the table of parameters above has the corresponding values of $x$ for each $\rho$.

Finally, this CAS code (Maple) ends the demonstration.

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UPDATE Nov.13 and Nov.21.2024. WZ PROOFS for Eqs. (1), (3), (5), (6), (7) and (8)

This approach was inspired by some comments of MO user @xiaoshuchong. We apply the WZ$_{s,t}$ transformation for a WZ pair $(F,G)$ pointed out in this MO question, namely to look for another WZ pair $(F_{s,t},G_{s,t})$ such that $F_{s,t}(n,k)=F(sn,k+tn)$ for some $s\in\mathbb{N},\,\,t\in\mathbb{Z}$ with $G_{s,t}(n,k)$ the resulting series pair we want to prove.

The base companion $F(n,k)$ is lifted up by a Beta function entanglement of two different linearly convergent logarithm series.

With $B(u,v)=\Gamma(u)\Gamma(v)/\Gamma(u+v)$ the Beta function, we use for $\log(2)$ $$F(n,k)=\frac38\cdot B(n+1,k+1/2)\cdot\frac{(\frac12)_n}{3^{2n}\,(1)_n}\cdot\frac{(-1)^k\,(1)_k}{2^{3k}\,(\frac12)_k}$$ as a seed to build $F_{s,t}(n,k)$. In fact, $$\sum_{k=0}^\infty F(0,k)=\frac{3}{4}\sum_{k=0}^\infty\frac{(-\frac12)^k}{(2k+1)\binom{2k}{k}}=\log(2)$$ The proofs of the corresponding series $\sum_{n=0}^\infty G_{s,t}(n,0)=\log(2)$ are obtained by creative telescoping algorithm certificates on these $F_{s,t}(n,k)$ terms.

We get Eqs.(3), (5), (6) and (7) proven by means of WZ$_{1,2}$, WZ$_{1,4}$, WZ$_{3,2}$ and WZ$_{3,4}$ respectively.

For $\log(3)$ we take the following companion,$$F(n,k)=\frac12\cdot B(n+1,k+1/2)\cdot\frac1{3^{n}}\cdot\frac{(-1)^k}{2^{2k}}$$ In fact,$$\sum_{k=0}^\infty F(0,k)=\sum_{k=0}^\infty \frac1{(2k+1)\,4^k}=\log(3)$$

Similarly, we get Eqs.(1) and (8) proven by means of WZ$_{2,1}$ and WZ$_{2,3}$ respectively.

Answer 3

As a supplement of @Jorge Zuniga's answer, we can obtain more similar series which are $\pi$ and $\log 2$ related via the same idea, such as: $$ \begin{eqnarray} \log2&=&\sum_{n=0}^{\infty}\frac{\left(\left(2n\right)!\right)^{2}\left(3n\right)!}{\left(n\right)!\left(6n+1\right)!}\left(\frac{1}{36}\right)^{n}\frac{598n+499}{144\left(6n+5\right)}\\\log2&=&\sum_{n=0}^{\infty}\frac{\left(\left(4n\right)!\right)^{2}}{\left(8n\right)!}\left(\frac{1}{324}\right)^{n}\frac{156128n^{3}+291728n^{2}+171658n+31441}{432\prod_{k=1}^{4}\left(8n+2k-1\right)}\\\pi&=&\frac{8}{3}\sum_{n=0}^{\infty}\frac{\left(\left(2n\right)!\right)^{2}\left(3n\right)!}{\left(n\right)!\left(6n+1\right)!}\left(-4\right)^{n}\frac{7n+6}{6n+5}\\\pi&=&\sum_{n=0}^{\infty}\frac{\left(\left(4n\right)!\right)^{2}}{\left(8n\right)!}4^{n+1}\frac{\left(3n+2\right)\left(112n^{2}+144n+41\right)}{\prod_{k=1}^{4}\left(8n+2k-1\right)} \end{eqnarray} $$ The more general case writes: $$ \begin{eqnarray} \alpha&=&\frac{\left(-1\right)^{l}b^{2l+2m}}{\left(b+1\right)^{l}\left(b+2\right)^{2m}}\\\sum_{k=1}^{l_{A}}A_{k}t^{k-1}&=&\frac{b+2}{b}\frac{1-\frac{\alpha t^{l}\left(1-t\right)^{m}}{4^{l}}}{t+\left(\frac{b+2}{b}\right)^{2}-1}\\\log\left(1+b\right)&=&f\left(\alpha,A\right)\\\beta&=&\frac{4^{l}}{\left(-1\right)^{m}\left(b^{2}+1\right)^{l}b^{2m}}\\\sum_{k=1}^{m+1}B_{k}t^{k-1}&=&-\frac{b}{2}\times\frac{1-\frac{\beta t^{l}\left(1-t\right)^{m}}{4}}{t-1-b^{2}}\\\arctan\frac{1}{b}&=&f\left(\beta,B\right) \end{eqnarray} $$ where $f$ writes $$ f\left(\alpha,A\right)=\sum_{n=0}^{\infty}\alpha^{n}\frac{\left(ln\right)!\left(\left(m+l\right)n\right)!\left(2mn\right)!}{\left(mn\right)!\left(2\left(m+l\right)n\right)!}\left[\sum_{k=1}^{m+1}\frac{A_{k}\Gamma\left(ln+k\right)}{\Gamma\left(ln+1\right)\prod_{j=1}^{k}\left(\left(m+l\right)n+\frac{2j-1}{2}\right)}\right] $$

One can simply describe the proof idea by only one step: The $\beta$ function connects the infinite series and the following two integrals $$ \begin{eqnarray} \log\left(1+b\right)&=&\frac{b+2}{b}\int_{0}^{1}\frac{1}{t+\left(\frac{b+2}{b}\right)^{2}-1}\frac{dt}{\sqrt{1-t}}\\\arctan\frac{1}{b}&=&-\frac{b}{2}\int_{0}^{1}\frac{1}{t-1-b^{2}}\frac{dt}{\sqrt{1-t}} \end{eqnarray} $$

Answer 4

This specific cubic integral is fascinating and has many nice properties

the original formula, including the kernel integral from which the series acceleration is built from, is found here https://arxiv.org/pdf/2402.08693

the $\ln(2)$ is found by

$15\sqrt{1 - w^2}×(\sin^{-1}(w))/w$ where $w = i/(2 \sqrt{2})$

using 2.10 and 2.9 in the aforementioned paper

$4/27 (w^6/(w^2-1))=-1/3888$ As shown in the original answer

To answer the question is if is known and proven, the answer is yes. The first examples of this cubic integral with the square root afik were published in 2008/2009 [1] to derive the pi formula, which follows from $z=1/48$.

$=\int_0^1\frac{P(x)x^{-1/2}}{1+zx(1-x)^2}dx$

The cubic without the square root is much older, which yields the Gosper formula. http://www.pi314.net/eng/schrogosper.php

[1] Wayback machine archive of original source shows the integral dates back to a at least a decade ago https://web.archive.org/web/20111125173313/http://www.iamned.com/math/ of the particular cubic integral: https://web.archive.org/web/20140623130232im_/http://upload.wikimedia.org/math/5/c/6/5c6c3251cd1a51bdf0780cf90ce4a49a.png.