I need to find what function is equal to the series $$\sum _{n=0}^{\infty }\frac{x^{4n}}{(2n)!} $$
I found the power series of $e^{x^2}$ and $e^{-x^2}$ but when I'm adding them together I don't understand why $n!$ should be changed to $2n!$, what the best technique for this type of questions?
$$e^{x^2} = \sum _{n=0}^{\infty }\frac{x^{2n}}{n!}$$
$$e^{-x^2} =\sum _{n=0}^{\infty }\frac{\left(-1\right)^nx^{2n}}{n!}$$
You are on the right track. If you sum the two series then you get $$ e^{x^2} + e^{-x^2}= \sum _{n=0}^{\infty }\frac{x^{2n}}{n!}+\sum _{n=0}^{\infty }\frac{\left(-1\right)^nx^{2n}}{n!} = \sum _{n=0}^{\infty } \bigl(1 + (-1)^n \bigr) \frac{x^{2n}}{n!} \, . $$ Now note that $1 + (-1)^n$ is zero for odd $n$ and equal to $2$ for even $n$, so that $$ e^{x^2} + e^{-x^2} = 2 \sum_{\substack{n=0 \\ \text{$n$ even}}}^{\infty } \frac{x^{2n}}{n!} \, . $$ Then substitute $n=2m$ in order to sum over all even indices: $$ 2 \sum_{\substack{n=0 \\ \text{$n$ even}}}^{\infty } \frac{x^{2n}}{n!} = 2 \sum_{m=0}^\infty \frac{x^{4m}}{(2m)!} \, . $$ Therefore $$ \sum_{m=0}^\infty \frac{x^{4m}}{(2m)!} = \frac 12 (e^{x^2} + e^{-x^2}) = \cosh(x^2) $$