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.

What you did wrong was changing $(2n)!$ to $2^nn!$.
You were correct that $e^x=\sum\limits_{n=0}^{\infty}\dfrac{x^{n}}{n!}$,
so $\cosh x = \dfrac{e^x+e^{-x}}2=\dfrac{\sum\limits_{n=0}^{\infty}\frac{x^{n}}{n!}+\sum\limits_{n=0}^{\infty}\frac{(-x)^{n}}{n!}}2=\dfrac{\sum\limits_{n=0}^{\infty}\frac{x^{n}}{n!}\left(1+(-1)^n\right) }2$.
$\dfrac{1+(-1)^n}2$ is $0$ when $n$ is odd and $1$ when $n$ is even, so this becomes $\sum\limits_{n=0}^{\infty}\dfrac{x^{2n}}{(2n)!} . $
