Proof of the product formula for \(\dfrac{\pi}{2\sqrt{3}}\)

 Hello everyone. Today we will prove the product formula for  \(\dfrac{\pi}{2\sqrt{3}}\). We will use the method of direct proof as a proof method. 

Theorem 1: \[\frac{\pi}{2\sqrt{3}}=\displaystyle\sum_{n=1}^{\infty}\frac{\chi(n)}{n}\]\[\text{where} \quad \chi(n)=\begin{cases} 1, & \text{if } n \equiv 1 \pmod{6}\\-1, & \text{if } n \equiv -1 \pmod{6}\\0, & \text{otherwise}\end{cases}\]

Theorem 2: We have\[\frac{\pi}{2\sqrt{3}}=\frac{5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdots}{6 \cdot 6 \cdot 12 \cdot 12 \cdot 18 \cdot 18 \cdot 24 \cdot 30 \cdots}\]expression whose numerators are the sequence of the odd prime numbers greater than \(3\) and whose denominators are even–even numbers one unit more or less than the corresponding numerators.


Proof:

By Theorem 1 we know that
\[\frac{\pi}{2\sqrt{3}}=1-\frac{1}{5}+\frac{1}{7}-\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+\frac{1}{19}-\cdots\]
we will have
\[\frac{1}{5} \cdot \frac{\pi}{2\sqrt{3}}=\frac{1}{5}-\frac{1}{25}+\frac{1}{35}-\frac{1}{55}+\cdots ,\]

and adding up both series,
\[\frac{6}{5} \cdot \frac{\pi}{2\sqrt{3}}=1+\frac{1}{7}-\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+\cdots .\]
We also have
\[\frac{1}{7} \cdot \frac{6}{5} \cdot \frac{\pi}{2\sqrt{3}}=\frac{1}{7}+\frac{1}{49}-\frac{1}{77}+\frac{1}{91}-\cdots ,\]

that subtracted from the former leads to
\[\frac{6}{7} \cdot \frac{6}{5} \cdot \frac{\pi}{2\sqrt{3}}=1-\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+\cdots ,\]
series where there are no denominators divisible either by 5 or by 7.

In a similar way we remove all those divisible by 11,
\[\frac{1}{11} \cdot \frac{6 \cdot 6}{7 \cdot 5} \cdot \frac{\pi}{2\sqrt{3}}=\frac{1}{11}-\frac{1}{121}+\frac{1}{143}-\cdots ;\]
that added to the former gives
\[\frac{12 \cdot 6 \cdot 6}{11\cdot 7 \cdot 5} \cdot \frac{\pi}{2\sqrt{3}}=1+\frac{1}{13}-\frac{1}{17}+\frac{1}{19}-\cdots .\]

We notice that the denominators which are divisible by a prime number of the form \(6n - 1\) are removed by addition, from which this new factor gets added \(\dfrac{6n}{6n-1}\) , while the denominators divisible by a prime number of the form \(6n + 1\) are removed by subtraction from which this new factor gets added, \(\dfrac{6n}{6n+1}\) .


Thus the denominators of these new factors successively added will be prime numbers while the numerators will be even–even numbers, one unit more or less than the denominators. Consequently if all the terms of the series considered at the beginning are subtracted in this way, we will finally have
\[\frac{\cdots 30 \cdot 24 \cdot 18 \cdot 18 \cdot 12 \cdot 12 \cdot 6 \cdot 6}{\cdots 29 \cdot 23 \cdot 19 \cdot 17 \cdot 13 \cdot 11 \cdot 7 \cdot 5} \cdot \frac{\pi}{2\sqrt{3}}=1 .\]

From this we will get
\[\frac{\pi}{2\sqrt{3}}=\frac{5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdots}{6 \cdot 6 \cdot 12 \cdot 12 \cdot 18 \cdot 18 \cdot 24 \cdot 30 \cdots}\]

\(\blacksquare\)



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