$f(x)=\sum_{n=0}^{+ \infty} \frac{(-1)^n}{(n!)^2}\left( \frac{x}{2}\right)^n $ is continuous 
Let \begin{align} f: \begin{cases} \mathbb{R} &\longrightarrow \mathbb{R} \\  x & \longmapsto \displaystyle \sum_{n=0}^{+ \infty} \frac{(-1)^n}{(n!)^2}\left( \frac{x}{2}\right)^n\end{cases} \end{align} 
  Show that:
(i) The function is well defined 
(ii) the function is continuous 

This is my first time I try to solve such a problem, especially (ii) is a concept I cannot grasp yet. 
For (i) I was told that I simply need to show that the function converges, if it wouldn't converge then it would not be considered as well defined. 
It is easy to show that the function converges using the d'Alembert Criteria (also known as the ratio test). For the sake of this question (and since it might be important later I will do this with some abbreviations).
\begin{align} \text{Let } u_n = \frac{(-1)^n}{(n!)^2}\left( \frac{x}{2}\right)^n, \ \text{such that } \frac{u_{n+1}}{u_n}=-\frac{1}{(n+1)^2} \left(\frac{x}{2} \right)  \\ \implies \left(\frac{x}{2} \right)\lim_{n \to + \infty} - \frac{1}{(n+1)^2}=0 \end{align}
So the function converges for all $x \in \mathbb{R}$ 
For (ii) I am stuck, to be brief about my problem is that I don't know about a sufficient way to show how a function is continuous. I know the definition of continuity at a given point $x_0$ and it is easy to grasp for me that if I can generalize this for all $x_0$ in the domain, then the function must be continuous. 
In Analysis 1 by Zorich continuous functions are demonstrated on some very easy examples such as $f(x)=x$ and some of the trigonometric functions. In the hints given by my tutor he recommends to use uniform convergence, but I don't see the link between uniform convergence and a function $f$ being continuous everywhere. 
Maybe if it would be possible to get me started on this problem with some basic steps, or trying to elaborate the intuition behind this problem, I could complete it on my own.
 A: (i) By ratio test, the series $\sum_{n=1}^\infty\frac{1}{(n!)^2}(\frac{|x|}{2})^n$ converges for any $x\in \mathbb{R}$. So $f$ is well-defined.
(ii) For any $x_0\in \mathbb{R}$, the series $\sum_{n=1}^\infty\frac{(-1)^n}{(n!)^2}(\frac{x}{2})^n$ converges on $[-|x_0|-1, |x_0|+1]$ uniformly. Since every $\frac{(-1)^n}{(n!)^2}(\frac{x}{2})^n$ is continuous on $x_0$, then $f$ is continuous on $x_0$. This shows that $f$ is continuous on $\mathbb{R}$.
A: Let's define $g_m(x)=\sum_{k=0}^m \frac{(-1)^k}{(k!)^2} \left( \frac{x}{2} \right)^k$. Of course $g_m$ is continuous, and it tends to $f$ pointwise. If we could show, that this convergence is uniform in neighbourhood of every point, it would imply that $f$ is continuous. This theorem is stated for example on wikipedia http://en.wikipedia.org/wiki/Uniform_convergence
'Uniform convergence theorem. If $(f_n)$ is a sequence of continuous functions which converges uniformly towards the function $ f$ on an interval $ S$, then $ f$ is continuous on $ S$as well.'
A: $$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{2^n(n!)^2}x^n=\sum_{n=0}^\infty a_nx^n \text{ where } a_n=\dfrac{(-1)^n}{2^n(n!)^2}~\forall~n\in\mathbb Z^+$$ 
Radius of convergence of the above power series is, $$\lim_{n\to\infty}\left|\dfrac{a_n}{a_{n+1}}\right|=\lim_{n\to\infty}|2(n+1)^2|=\infty$$
A power series being convergent within its circle of convergence and being continuous thereon (i) and (ii) follow.
