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
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Proof of the theorem regarding Fibonacci prime numbers

Hello everyone. Today we will prove the theorem regarding Fibonacci prime numbers. We will use the method of contradiction as a proof method. 

Theorem: Let \(F_n \) be the nth Fibonacci number and let \(F_n \) be a prime number. Then \(n \) is also a prime number, except in the case of \(F_4 = 3 \). 

Proof: 

For the case when \(n = 2 \) we have that \(F_2 = 1 \), and as we know the number \(1 \) is neither simple nor composite, so this case does not refute the truth of the theorem. 

For the case when \(n = 3 \) we have that \(F_3 = 2 \), so this case is in accordance with the statement. 

Now, suppose that for \(n> 4 \), \(F_n \) is a prime number and that \(n = rs \) for some natural numbers \(r, s \) which are greater than \(1 \), that is, that \(n \) is a composite number. 

Since \(n> 4 \)  at least one of the numbers \(r \) and \(s \) is greater than \(2 \). Then, according to the divisibility theorem of Fibonacci numbers which reads: \[\forall m, n \in \mathbb {Z} _ {> 2} \quad m \mid n \Leftrightarrow F_m \mid F_n \] we have that at least one of the expressions \(F_r \mid F_n \) and \(F_s \mid F_n \) is correct. Further, since at least one of the numbers \(r \) and \(s \) is greater than \(2 \), it follows that at least one of the Fibonacci numbers \(F_r \) and \(F_s \) is greater than \(1 \). So, we have shown that the number \(F_n \) has at least one divisor greater than \(1 \) and less than \(F_n \), i.e. we have shown that \(F_n \) is a composite number, thus we arrived at a contradiction. From this we conclude that \(n \) must be a prime number. 

\(\blacksquare \) 

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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
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