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 formula for the sum of the first n natural numbers

Hello everyone. Today we will prove the formula for the sum of the first \(n \) natural numbers. We will use the method of mathematical induction as a proof method. It is believed that the Pythagoreans also knew this formula. 

Theorem: For any natural number \(n \) the following formula holds: \[\displaystyle \sum_{k = 1} ^ nk = \frac {n (n + 1)} {2} \] Proof: 

1. Base case (n = 1) \[\displaystyle \sum_{k = 1} ^ {1} k = \frac {1 \cdot (1 + 1)} {2} \] \[1 = \frac {1 \cdot (2)} {2} \] \[1 = \frac {2} {2} \] \[1 = 1 \] 

2. Induction hypothesis (n = m) 

Suppose that: \[\displaystyle \sum_{ k = 1} ^ mk = \frac {m (m + 1)} {2} \] 

3. Inductive step (n = m + 1) 

Using the assumption from the second step, we will prove that: \[\displaystyle \sum_{k = 1} ^ {m + 1} k = \frac {(m + 1) (m + 2)} {2} \] So, \[\displaystyle \sum_{k = 1} ^ {m + 1} k = \displaystyle \sum_{k = 1} ^ {m} k + m + 1 = \] \[\frac {m (m +1)} {2} + m + 1 = \] \[\frac {m ^ 2 + m + 2m + 2} {2} = \] \[\frac {m ^ 2 + 2m + m + 2} {2} = \] \[\frac {m (m + 2) + (m + 2)} {2} = \] \[\frac {(m + 1) (m + 2)} {2} \]

 \(\blacksquare \)

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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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