Proof that I0(2)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.

$\blacksquare$