Please note that this is a "follow up" of my previous question $f(x)=\sum_{n=0}^{+ \infty} \frac{(-1)^n}{(n!)^2}\left( \frac{x}{2}\right)^n $ is continuous. I decided to accept the answer given there because under the circumstances I was asking the question the answers I got were sufficient. I now have the possibility to refine the question and show where I need some help.
Show that $$f(x)=\sum_{n=0}^{+\infty} \frac{(-1)^n}{(n!)^2}\left( \frac{x}{2} \right)^n$$ converges uniformly on the interval $ [x_0-1, x_0+1]$
Introduction: First allow me to highlight that we did not cover the Weierstrass-M-Test (Criteria) for uniform convergence, from what I have read from various online resources (including exercise videos) the Weierstrass-M-Test seems to be one of the strongest tools to check for uniform convergence.
Question: Is it possible to show that the Bessel Function converges uniformly on $[x_0-1,x_0+1]$ by using the same technique as in the example given.
Example: Allow me to demonstrate the way we did this in class with examples I do understand (the examples are simple, the tough ones are left for us but 'should' be possible to be handled in the same way).
$f_n(x) = \frac{x}{n}$ where $x \in [0,1]$ such that $f_{n}(x)=0$ for $n \to \infty$ (point wise convergence)
$|f_n(x)-0|= \left| \frac{x}{n} \right|= \frac{x}{n}\leq \frac{1}{n} \to 0$ for $ n \to \infty$ (uniform convergence on the intervall)
Counterexample:
$$f_n(x)= \frac{n^2x}{1+(nx)^4}, \ x \in [ 0,1] \implies \lim_{n \to \infty} f_n(x)=0 \tag{point wise }$$ Let $x_n= \frac{1}{n}$ $$ \implies f_n(x_n)=\frac{n}{2}\implies |f_n(x_n)-0|= \frac{n}{2}> \epsilon \\ \implies \text{ not uniform convergence}$$
Back to the Bessel Function: It is easy to show that the Bessel Function converges using the d'Alembert Criteria (Ratio Test). However, I cannot say to which value the Bessel Function converges, which is crucial for the method introduced above. Furthermore, for each $n \in \mathbb{N}$, $f_n(x)$ the Bessel Function is a polynomial and therefore continuous, especially continuous on a given $x_0$. And here I am stuck.
With Weierstrass-M: Supplementary for this question but I will still add my approach:
fix $x_0=2$ such that the Intervall becomes $[1,3]$ then $$\frac{1}{(n!)^2}\left( \frac{x}{2}\right)^n \leq \frac{1}{(n!)^2}\left(\frac{1}{2} \right)^n=:M_n$$ which seems to go wrong for n approaching infinity the given $M$ would tend to zero and fall out of the given interval.