Series for $1/\pi$ arising from Cauchy product

This paper proves a conjectured series for $1/\pi$ by Sun using Cauchy products and hypergeometric transformations, and subsequently derives two additional analogous series involving cubic polynomials along with a table of further identities.

The Gist

Imagine you are trying to measure the circumference of a perfect circle. For thousands of years, mathematicians have been obsessed with finding the exact value of $\pi$ (pi). But here's a twist: instead of trying to calculate $\pi$ directly, some modern mathematicians are trying to calculate $1/\pi$ (one divided by pi).

Why? Because $1/\pi$ is like a secret code. If you can write down a formula that adds up to exactly $1/\pi$, you can use that formula to calculate $\pi$ to millions of decimal places very quickly.

This paper is about cracking a specific, very difficult code in that secret language.

The Puzzle: A Giant Math Jigsaw

The author, Roman Le Lan, is solving a puzzle left by a mathematician named Zhi-Wei Sun. Sun had a list of 37 "conjectures" (educated guesses) about how to build $1/\pi$ using a specific type of mathematical recipe.

Think of these recipes as Lego sets.

• The basic blocks are numbers and fractions.
• The instruction manual involves a complex stacking pattern called a "sum" (adding up an infinite list of numbers).
• Sun guessed that if you stack these blocks in a very specific, weird way, the final tower would equal exactly $1/\pi$.

Sun's team had already solved 36 of these puzzles. The last one (Theorem 1 in the paper) was the "final boss" that no one had beaten yet.

The Magic Trick: The Cauchy Product

How did Le Lan solve the final boss? He used a tool called the Cauchy Product.

Imagine you have two long lines of people, each holding a sign with a number on it.

1. Line A is holding signs with a specific pattern.
2. Line B is holding signs with a different pattern.

Usually, you just add the numbers in Line A together. But the Cauchy Product is like a dance. You take the first person from Line A and pair them with the last person from Line B, then the second from A with the second-to-last from B, and so on. You multiply their numbers and add them all up.

In this paper, Le Lan took two complex mathematical "lines" (called hypergeometric series) and made them dance together. When they paired up, they didn't just make a mess; they transformed into a simpler, cleaner pattern that he recognized from a famous formula discovered by the legendary mathematician Srinivasa Ramanujan.

It's like taking two complicated, tangled headphones, twisting them together in a specific way, and suddenly they untangle into a perfect, straight wire.

The Results: Cracking the Code

Once he untangled the knot, Le Lan proved that Sun's final guess was correct. He showed that this specific, messy-looking sum of numbers actually equals:
$$\frac{3\sqrt{6}}{2\pi}$$
(Which means if you rearrange the math, you get a perfect value for $1/\pi$).

But he didn't stop there. Because he understood the "dance" (the method), he could teach the lines to dance in slightly different ways.

• Theorem 2: He used the same dance steps but changed the music (the numbers in the recipe) to create two new formulas for $1/\pi$. These new formulas are even more complex, involving cubes of numbers ($n^3$), but they work perfectly.

The "Unsolvable" Mystery

There is one part of the paper the author admits he couldn't solve. Sun also guessed that if you stop the infinite sum early (at a prime number $p$), the result follows a specific rule in "modular arithmetic" (a kind of clock math).

Le Lan says, "I proved the infinite version works perfectly, but I couldn't prove the 'clock math' version." It's like proving a bridge is strong enough to hold a truck, but not yet proving it won't shake if a specific type of wind blows. That remains a mystery for someone else to solve.

Why Does This Matter?

You might ask, "Who cares about these weird sums?"

1. Computing Power: These formulas are incredibly efficient. Computers use them to calculate $\pi$ to record-breaking lengths.
2. Mathematical Beauty: It shows that the universe of numbers is connected in surprising ways. A messy sum of fractions involving $-1/3$ and $-1/6$ somehow connects to the circle constant $\pi$.
3. New Tools: Le Lan didn't just solve one problem; he gave us a new "key" (the method) that can unlock many other similar doors. He even lists 8 more potential formulas at the end of the paper that other mathematicians can now try to prove.

In short: This paper is a masterclass in taking a messy, unsolved math puzzle, using a clever "dance" technique to simplify it, and revealing a beautiful, hidden connection to the number $\pi$.

Technical Summary

1. Problem Statement

The paper addresses the evaluation of specific infinite series of the form:
$$\sum_{n=0}^{\infty} (an + b)q^n \sum_{k=0}^{n} \binom{-x}{k} \binom{x-1}{n-k} \binom{-y}{k} \binom{y-1}{n-k} = \frac{\alpha}{\pi}$$
where $a, b$ are nonzero integers, $q, x, y$ are nonzero rational numbers, and $\alpha$ is an algebraic number of degree at most 2.

These series are part of the broader class of Ramanujan-Sato type series for $1/\pi$. The specific problem tackled is the proof of the final unproven identity from a list of 37 conjectures proposed by Zhi-Wei Sun. While Almkvist and Aycock had previously evaluated 36 of these, the specific case involving parameters $x=1/3, y=1/6$ (leading to binomial coefficients with arguments $-1/3, -2/3, -1/6, -5/6$) remained open.

2. Methodology

The author employs a combination of hypergeometric transformations, the Cauchy product of series, and Zeilberger's algorithm (for recurrence relations). The methodology proceeds in three main stages:

A. Transformation via Cauchy Product

The core of the proof involves defining a generating function $P(z)$ as the product of two hypergeometric series:
$$P(z) = \left( \sum_{k=0}^{\infty} z^k \binom{-1/3}{k} \binom{-1/6}{k} \right) \left( \sum_{k=0}^{\infty} z^k \binom{-2/3}{k} \binom{-5/6}{k} \right)$$
Using the Cauchy product rule, this is rewritten as a single series involving a convolution of binomial coefficients. The author then utilizes Euler's transformation to express $P(z)$ as a product of two ${}_2F_1$ hypergeometric functions:
$$P(z) = {}_2F_1\left(\frac{1}{3}, \frac{1}{6}; 1; z\right) {}_2F_1\left(\frac{2}{3}, \frac{5}{6}; 1; z\right) = (1-z)^{-1/2} \left[ {}_2F_1\left(\frac{1}{3}, \frac{1}{6}; 1; z\right) \right]^2$$

B. Lemma 1: Combinatorial Identity

A crucial intermediate step (Lemma 1) establishes a transformation for the coefficients of the squared hypergeometric series. Using Zeilberger's algorithm, the author proves the identity:
$$108^n \sum_{k=0}^{n} \binom{-1/3}{k} \binom{-1/6}{k} \binom{-1/3}{n-k} \binom{-1/6}{n-k} = \binom{2n}{n}^2 \binom{3n}{n}$$
This allows the generating function $P(z)$ to be expressed in terms of a known series involving central binomial coefficients and the term $(1-z)^{-1/2}$.

C. Operator Application and Differentiation

To extract the specific linear term $(3n-1)$ required for the conjecture, the author applies a differential operator to the generating function evaluated at a specific point ($z = 1/2$):
$$\left( -1 + \frac{3}{2}z \frac{d}{dz} \right) \bigg|_{z=1/2}$$
This operation transforms the series into a form that matches a known Ramanujan-type series for $1/\pi$ provided by J. Guillera (Lemma 2).

D. Proof of Theorem 2: Recurrence Relations

For the second set of results (involving cubic polynomials in $n$), the author uses Zeilberger's algorithm directly on the sequence $a_n$ (the inner sum of the series). This yields a linear recurrence relation of order 2. By summing this recurrence over $n$ and manipulating the resulting telescoping sums, the author derives new identities involving polynomials of degree 3 ($9n^3 - \dots$ and $18n^3 - \dots$).

3. Key Contributions and Results

Theorem 1: Proof of Sun's Conjecture

The paper proves the specific identity conjectured by Sun:
$$\sum_{n=0}^{\infty} \frac{3n-1}{2^n} \sum_{k=0}^{n} \binom{-1/3}{k} \binom{-2/3}{n-k} \binom{-1/6}{k} \binom{-5/6}{n-k} = \frac{3\sqrt{6}}{2\pi}$$
This resolves the last remaining case of Sun's 37 conjectures regarding this specific form of series.

Theorem 2: New Cubic Series

Extending the method, the author derives two new series involving cubic polynomials in the numerator:

1. $$\sum_{n=0}^{\infty} \frac{9n^3 - 27n^2 - 14n - 2}{2^n} (\dots) = -\frac{3\sqrt{6}}{8\pi}$$
2. $$\sum_{n=0}^{\infty} \frac{18n^3 - 54n^2 - 25n - 5}{2^n} (\dots) = \frac{3\sqrt{6}}{4\pi}$$

Table of Additional Identities

The paper notes that the same methodology can be applied to prove eight additional series of the form $(an+b)q^n \dots = \alpha/\pi$, which are listed in a table at the end of the note.

Unresolved Conjecture

The paper mentions a supercongruence conjecture by Sun related to the main identity modulo $p^2$ for primes $p > 3$. The author explicitly states that this congruence was not proven in this work.

4. Significance

• Completion of a Conjecture List: The primary significance is the closure of Sun's list of 37 conjectures for this specific class of series, providing a rigorous proof for the final case.
• Methodological Efficiency: The paper demonstrates a powerful, unified approach using the Cauchy product of hypergeometric series combined with differential operators. This method avoids the need for complex modular form theory often used in Ramanujan-Sato proofs, relying instead on classical hypergeometric identities and algorithmic recurrence finding.
• Expansion of Known Series: By deriving Theorem 2, the paper expands the known landscape of $1/\pi$ series to include those with cubic polynomial numerators, which are generally more difficult to derive than linear ones.
• Foundation for Future Work: The technique presented serves as a template for proving further identities, as evidenced by the additional results listed in the paper's table.

In summary, Roman Le Lan provides a concise and elegant proof for a significant open problem in number theory, utilizing a blend of combinatorial identities and hypergeometric calculus to generate new series for $1/\pi$.