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 \(I_0(\sqrt{2})\) is an irrational number

  Hello everyone. Today we will prove that  \(I_0(\sqrt{2})\) is an irrational number, where \( I_0\) denotes a modified Bessel function of the first kind. We will use the method of contradiction as a proof method.

Theorem 1:  \(I_0(\sqrt{2})=\displaystyle\sum_{n=0}^{\infty} \frac{1}{(n!)^22^n}\)

Theorem 2:  \(I_0(\sqrt{2})\)  is an irrational number.

Proof:

Suppose that  \(I_0(\sqrt{2})\)  is a rational number. Then  \(I_0(\sqrt{2})\)  can be written in the form \( \dfrac{p}{q} \) where \( p \)  and \( q\) are coprime integers such that \(q \ge 1\). By Theorem 1 we can write the following equality: \(q!(q-1)!p2^q=\displaystyle\sum_{n=0}^{\infty} \frac{(q!)^22^q}{(n!)^22^n}\) . Since left hand side of this equality is an integer the sum \(\displaystyle\sum_{n=q+1}^{\infty} \frac{(q!)^22^q}{(n!)^22^n}\) which is greater than zero \(0\)  also must be an integer. Clearly for \(n \ge q+1\) we have , \(\left(\dfrac{q!}{n!}\right)^2 \le \dfrac{1}{(q+1)^{2(n-q)}}\) , hence \(\displaystyle\sum_{n=q+1}^{\infty} \frac{(q!)^22^q}{(n!)^22^n}\le \displaystyle\sum_{n=q+1}^{\infty} \frac{1}{(q+1)^{2(n-q)}\sqrt{2}^{2(n-q)}}=\displaystyle\sum_{k=1}^{\infty} \frac{1}{\left(2(q+1)^2\right)^k}=\frac{1}{2(q+1)^2-1}<1\) . So, we have shown that the following inequalities hold true: \(0< \displaystyle\sum_{n=q+1}^{\infty} \frac{(q!)^22^q}{(n!)^22^n}<1\) from which it follows that: \(\displaystyle\sum_{n=0}^{\infty} \frac{(q!)^22^q}{(n!)^22^n} \not\in\mathbb{Z}\)  , So. we arrived at a contradiction, hence   \(I_0(\sqrt{2})\)  is an irrational 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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