Show that the following function is continuous on the set of real numbers: $f(x)=\sum _{\ n=0}^{\infty} (\frac{x^n}{n!})^2$ I think that I will have to use the ratio test, but I am not too sure. 
 A: Fix a bound $M$ and consider the closed interval $[-M, M]$. Note that $$\left|\left(\frac{x^n}{n!}\right)^2\right| \le \frac{M^{2n}}{(n!)^2}$$
So we study convergence of the series
$$\sum_{n = 0}^{\infty} \frac{M^{2n}}{(n!)^2}$$
If it converges, the Weierstrass $M$-test gives that the desired summation is continuous. Note that after some algebra,
\begin{align*}
\lim_{n \to \infty} \frac{M^{2(n + 1)}}{((n + 1)!)^2} \cdot \frac{(n!)^2}{M^{2n}} &= M \left(\lim_{n \to \infty} \frac{n!}{(n + 1)!}\right)^2 \\
&= M \left( \lim_{n \to \infty} \frac 1 {n + 1}\right)^2 \\
&= 0
\end{align*}
Can you put this all together?
A: If you write it as $\sum_n \frac{(x^2)^n}{n!^2}$ you can see that it's a power series in $x^2$. A power series is continuous in every open interval where it converges (*), so all we need to show is that the series converges everywhere.
But that is easy -- each term is positive and less than the corresponding term in the power series for $e^{x^2}$ by a factor of $n!$, so it converges by the comparison test.
(*) I'm not sure if this fact counts as real analysis -- I think it's usually taught as part of complex analysis, but it's equally true on the real line.
