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Recently, I discovered the following identity

$$\sum_{k=1}^{\infty}\frac{100+299 k}{\left(24 k^{2}+8 k \right) 576^{k}{\binom{3 k}{k}} }=\frac{1}{8}+3\log(2)-2\log(3). \tag{1}$$

Based on identity (3) in this MO Question, the series (1) provides a fast algorithm for $\log(3)$.

My questions are:

  1. Is the series (1) known? Can it also be proven using the WZ method?

  2. Is this the fastest algorithm for calculating $\log(3)$ currently?

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  • $\begingroup$ @Deyi_Chen Interesting unknown series having same $\rho=\frac1{3888}$ but different hypergeometric motive that is now $[[1\,\frac12],[\frac13\,\frac23]]$ instead of the former $[[1\,\frac12],[\frac16\,\frac56]]$. For speed all is about costs. This identity has this $\log\,3$ cost = 0.968 ($\log\,2$) + 0.968 (series) = 1.936. It does not compare to Eq.(1) in that MO link with cost 1.434 nor Eq.(8) with cost 1.646. Also, if an hypergeometric series for logarithms has a cost over 1.95, the Machin type formula becomes faster. $\endgroup$ Commented Jan 29 at 17:29

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Note that $$\frac1{3k(3k+1)}=\frac1{3k}-\frac1{3k+1}$$ and that $100+299k = 100(1+3k)-k$. So $$\frac{100+299k}{24k^2+8k}=\frac{100}{8k}-\frac1{8(3k+1)}.$$ Therefore the problem reduces to the evaluations of the sums $$\sum_{k=1}^\infty\frac{x^k}{k^r\binom{3k}k}\ \text{and}\ \sum_{k=0}^\infty \frac{x^k}{(3k+1)\binom{3k}k}$$ with $x=1/576$ and $r\in\{0,\pm1\}$. These sums with $|x|<27/4$ have been evaluated, see this paper, this paper, arXiv:2401.12083, and Batir's papers cited there. For $|x|<27/4$, we also have $$\sum_{k=1}^n((4x-27)k+2x+27)\frac{x^k}{\binom{3k}k}-6\sum_{k=1}^n\frac{x^k}{k\binom{3k}k}=-2x+\frac{(4n+2)x^{n+1}}{\binom{3n}n}$$ which tends to $-2x$ as $n\to\infty$.

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  • $\begingroup$ Thank you for your excellent answer! $\endgroup$ Commented Jan 30 at 7:23
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I will expand my comment above. This identity can be written in this hypergeometric form, which allows to evaluate the speed or efficiency of the formula$$\log\left(\frac98\right)=\frac12\cdot\sum_{n=1}^\infty\left(\frac1{2^4\,3^5}\right)^n\cdot\frac{578\,n-191}{n(2n-1)}\cdot\left[\begin{matrix} 1&\frac{1}{2}\\ \frac{1}{3}&\frac{2}{3}\\ \end{matrix}\right]_n$$ You can evaluate the efficiency for this and other series of the same kind by computing the binary splitting cost, where $\rho$ is the convergence rate and $d$ is the number of Pochhammers in a row. $$C_s=-4\cdot\frac{d}{\log(\rho)}$$ A series is faster if it has a lower cost. Formulas having different series or constants must add their costs. I have already explained it here in MO in other Q&A's. Now, answering your questions,

Q1a. Formula is not known explicitly, but as it is deduced from the answer by Zhi-Wei Sun, it can be proven easily by a linear combination of the indicated sums whose closed forms are found in the references given.

Q1b. WZ proof. We need a proper (F,G) pair from the series known. This is not easy in this case. The method of Au's seeds fails, and we cannot use this tool. This is a very similar case to my Question about some logarithm formulas. I could find (after some months) an entanglement of two linearly convergent logarithm series providing the searched (F,G) pair that allowed to prove them and discover more proven identities.

Q2. About the speed. No, this is not the fastest formula for $\log(3)$. The fastest is Eq.(1) here having $C_s=1.464..$ and the 2nd fastest is Eq.(8) with $C_s=1.646..$. For $\log(2)$ the fastest has $C_s=0.968..$.

Your identity (1) has a series cost $$C_{series}=-4\cdot\frac{2}{\log\left(\frac1{3888}\right)}=0.968..$$

but you have to add the $C_{\log(2)}=0.968..$ which gives a total cost for $\log(3)$ equal to $C_s=1.936..$ that is worse than $1.464$.

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  • $\begingroup$ Thank you for your excellent answer. It seems that this series provides a relatively fast algorithm for calculating $\log(9/8)$ (with $C_s=0.968..$). In other words, when $n=1$ and $n=8$, there are algorithms that make the calculation of $\log(1+\frac{1}{n})$ have a speed of $C_s=0.968..$. Exploring whether there are similarly efficient algorithms for other values of $n$ also seems to be an interesting problem. $\endgroup$ Commented Jan 30 at 7:34
  • $\begingroup$ Is the series I discovered the fastest algorithm currently for calculating both $\log(2)$ and $\log(3)$ simultaneously (with a total $C_s=1.936..$)? $\endgroup$ Commented Jan 31 at 10:55
  • $\begingroup$ It is just the fastest for $\log(9/8)$ with a cost 0.968 and nothing more. $\endgroup$ Commented Jan 31 at 11:27
  • $\begingroup$ Currently, the fastest computation for $\log(2)$ is $0.968$, and for $\log(3)$ it is $1.464$. Does this mean that if we compute $\log(2)$ and $\log(3)$ separately to obtain their precise values, it would take $0.968 + 1.464 = 2.432$, which is slower than $1.936$? $\endgroup$ Commented Jan 31 at 11:39
  • $\begingroup$ @Deyi_Chen No, $C_s=1.464$ is obtained for $\log(3)$ straightwise with its own formula from Eq.(1) in my MO Post (the link in my answer) with no $\log(2)$ participation. This is the Cost you have to beat with a new (mixed or not) formula for $\log(3)$ $\endgroup$ Commented Jan 31 at 13:48

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