Prove some properties of the Bessel functions of the first kind of orders zero and one

We define the Bessel functions of the first kind of orders zero and one by

$J_0 (x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(n!)^2 2^{2n}}, \qquad J_1 (x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{n!(n+1)! 2^{2n+1}}.$

Prove that both $J_0(x)$ and $J_1(x)$ converge for all $x \in \mathbb{R}$ .
Prove that $J'_0(x) = -J_1(x)$ .
If we define two new functions
$j_0(x) = xJ_0(x), \qquad j_1 (x) = x J_1(x)$

prove that $j_0(x) = j_1'(x)$ .

Proof. For the order zero Bessel function of the first kind $J_0$ we have $a_n = \frac{x^{2n}}{(n!)^2 2^{2n}}$ and so using the ratio test we have
$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{x^{2n+2}}{(n+1)!^2 2^{2n+2}} \right) \left( \frac{(n!)^2 2^{2n}}{x^{2n}} \right) \\[9pt] &= \lim_{n \to \infty} \frac{x^2}{4 (n+1)^2} \\[9pt] &= 0. \end{align*}$

Hence, $J_0 (x)$ converges for all $x \in \mathbb{R}$ .

For $J_1(x)$ we have $a_n = \frac{x^{2n+1}}{n!(n+1)! 2^{2n+1}}$ and so

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{x^{2n+3}}{(n+1)! (n+2)! 2^{2n+3}} \right) \left( \frac{n!(n+1)! 2^{2n+1}}{x^{2n+1}} \right) \\[9pt] &= \lim_{n \to \infty} \frac{x^2}{4(n+1)(n+2)} \\[9pt] &= 0. \end{align*}$

Hence, $J_1(x)$ converges for all $x \in \mathbb{R}. \qquad \blacksquare$
Proof. We compute the derivative of $J_0(x)$ directly,
$\begin{align*} && J_0(x) &= \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(n!)^2 2^{2n}} \\[9pt] \implies && J_0'(x) &= \sum_{n=1}^{\infty} (-1)^n \frac{2nx^{2n-1}}{(n!)^2 2^{2n}} \\[9pt] &&&= \sum_{n=1}^{\infty} (-1)^n \frac{x^{2n-1}}{(n-1)!n! 2^{2n-1}} \\[9pt] &&&= \sum_{n=0}^{\infty} (-1)^{n+1} \frac{x^{2(n+1)-1}}{(n+1-1)!(n+1)! 2^{2(n+1)-1}} \\[9pt] &&&= - \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{n! (n+1)! 2^{2n+1}} \\[9pt] &&&= - J_1(x). \qquad \blacksquare \end{align*}$
Proof. First, we have
$j_0 (x) = x J_0(x) = x \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(n!)^2 2^{2n}} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(n!)^2 2^{2n}}.$

On the other hand we have,

$j_1 (x) = xJ_1(x) = x \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{n!(n+1)! 2^{2n+1}} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+2}}{n!(n+1)! 2^{2n+1}}.$

Therefore,

$\begin{align*} j_1'(x) &= \sum_{n=0}^{\infty} (-1)^n \frac{(2n+2)x^{2n+1}}{n! (n+1)! 2^{2n+1}} \\[9pt] &= \sum_{n=0}^{\infty} (-1)^n \frac{2(n+1)x^{2n+1}}{n! (n+1)! 2^{2n+1}} \\[9pt] &= \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(n!)^2 2^{2n}}. \end{align*}$

Hence, we indeed have $j_0(x) = j_1'(x). \qquad \blacksquare$