Find the sum of series: $\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$ I have some trouble with series theory. The specific questions are as follows:
\begin{equation}
\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}
\end{equation}
My idea is just like this:
Since $e^x=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$,
\begin{align}
\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}&=\sum_{n=0}^{\infty}\frac{x^{2n}}{2^nn!}\\
&=\sum_{n=0}^{\infty}\frac{(\frac{x^2}{2})^n}{n!}\\
&=e^{\frac{x^2}{2}}
\end{align}
However, the answer is cosh $x$. The main idea is based on the power series of $e^x$ and $e^{–x}$. Then add them together. But I still don't understand what I did wrong.
Can anyone help me out,please. Thank you.

$$\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} $$
$$ =\frac{1}{2} \sum_{n=0}^{\infty}  \frac{2 \cdot x^{2n}}{(2n)!} -\frac{1}{2} \sum_{n=0}^{\infty}  \frac{x^{2n+1}}{(2n+1)!} + \frac{1}{2} \sum_{n=0}^{\infty}  \frac{ x^{2n+1}}{(2n+1)!} $$
$$ =\frac{1}{2} \sum_{n=0}^{\infty}  \frac{x^{n}}{(n)!} + \frac{1}{2} \sum_{n=0}^{\infty}  \frac{(-x)^{n}}{(n)!} $$
$$ = \frac{e^{x}+e^{-x}}{2} $$
$$= cosh(x) $$
