Question: Prove that the function $f(x)$ defined as $$f(x)=\sum_{n=0}^\infty\left(\frac{x^n}{n!}\right)^2$$ is continuous on $\mathbb{R}$, for any real number $x$.
My Attempt: Given $\epsilon$ and a fixed real number $x_0$, choose $$\delta(\epsilon, x_0)=\frac{\epsilon}{\left|\sum_{n=0}^\infty\frac{x_0^{n-1}\times 2x_0^n}{(n-1)!n!}\right|+1}$$ Then, $$|x-x_0|<\delta \\\Rightarrow |x-x_0|\left|\sum_{n=0}^\infty\frac{x_0^{n-1}\times 2x_0^n}{(n-1)!n!}\right|<\frac{\epsilon}{\left|\sum_{n=0}^\infty\frac{x_0^{n-1}\times 2x_0^n}{(n-1)!n!}\right|+1}\times\left|\sum_{n=0}^\infty\frac{x_0^{n-1}\times 2x_0^n}{(n-1)!n!}\right|<\epsilon \\\Rightarrow \left|\sum_{n=0}^\infty\left(\left(\frac{x^n}{n!}\right)^2-\left(\frac{x_0^n}{n!}\right)\right)^2\right|<\epsilon \\\Rightarrow |f(x)-f(x_0)|<\epsilon$$ Hence, $f(x)$ is continuous for $x_0$. But since $x_0$ was arbitrary. Hence, $f(x)$ is continuous for $\mathbb{R}$
Any mistakes in my proof? Please correct me. Also, if there is any other way to prove $f(x)$ is continuous? Thanks in advance.
Hint: The following theorem is well known Theorem : Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of functions with common domain $X$ and suppose that $f_n$ is continuous at $c \in X$ for all $n \in \mathbb{N}$. If $\sum_{n=1}^{\infty} f_n(x)$ uniformly converges to the sum function $f(x)$ then $f$ is continuous at $c$.
All you need to check is that your series is uniformly convergent.