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 that the neighboring Fibonacci numbers are coprime

Hello everyone. Today we will prove that the neighboring Fibonacci numbers are coprime. We will use the method of mathematical induction as a proof method. 

Theorem: Let \(F_n \) represent the nth Fibonacci number. Then: \[\forall n \ge 2, \quad \operatorname{NZD} \left (F_n, F_ {n + 1} \right) = 1 \] 

Proof: 

1. Base case (n = 2) 

\[ \operatorname{NZD} \left(F_2, F_{3} \right) = \operatorname{NZD} (1,2) = 1 \] 

2. Induction hypothesis (n = m) 

Suppose that: \[\operatorname { NZD} \left (F_m, F_ {m + 1} \right) = 1 \] 

3. Inductive step (n = m + 1) 

Using the assumption from the second step, we will prove that: \[\operatorname{NZD}\left(F_{m+1},F_{m+2}\right)=1\]Since the greatest common divisor of any natural numbers \(a \) and \(b \) is equal to the greatest common divisor of any linear combination of numbers \(a \) and \( b \) we have that \(\operatorname {NZD} (a, b) = \operatorname {NZD} (a, ba) \). With this in mind, we can write the following equation: \[\operatorname {NZD} \left (F_ {m + 1}, F_ {m + 2} \right) = \operatorname {NZD} \left(F_ {m + 1} , F_ {m + 2} -F_ {m + 1} \right) \] Next,  by the definition of the Fibonacci sequence \(F_ {m + 2} = F_ {m} + F_ {m + 1} \) we have that: \[\operatorname {NZD} \left (F_ {m + 1}, F_ {m + 2} \right) = \operatorname {NZD} \left (F_ {m + 1}, F_ {m} \right) \] \[\operatorname {NZD} \left (F_ {m + 1}, F_ {m + 2} \right) = \operatorname {NZD} \left (F_ {m}, F_ {m + 1} \right) = 1 \] 

\(\blacksquare \)

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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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