Find the sum of $\sum\limits_{n=1}^{\infty} \frac{x^{2n}}{(2n)!}$ Find the sum of $\sum_{n=1}^{\infty} \dfrac{x^{2n}}{(2n)!}$  on its interval of convergence.
We can see that the domain of convergence is $D=R$. Then let:
$$f(x)=\sum_{n=1}^{\infty} \dfrac{x^{2n}}{(2n)!}$$
$$f'(x)=\sum_{n=1}^{\infty} \dfrac{x^{2n-1}}{(2n-1)!}$$
$$f''(x)=\sum_{n=1}^{\infty} \dfrac{x^{2n-2}}{(2n-2)!}$$
Thus $f''(x)=f(x)$, solve this differential equation, we'll get the solution.
Is my solution right? I just begin to study the power series. Thank you so much.

Let $\omega=\exp\left(\pi i\right)
 $. Then $$e^{\omega x}=\sum_{n\geq0}\frac{x^{n}\omega^{n}}{n!}=\sum_{n\geq0}\frac{x^{2n}}{\left(2n\right)!}-\sum_{n\geq0}\frac{x^{2n+1}}{\left(2n+1\right)!}
 $$ and $$e^{-\omega x}=\sum_{n\geq0}\frac{x^{n}\left(-\omega\right)^{n}}{n!}=\sum_{n\geq0}\frac{x^{2n}}{\left(2n\right)!}+\sum_{n\geq0}\frac{x^{2n+1}}{\left(2n+1\right)!}
 $$ hence $$\sum_{n\geq0}\frac{x^{2n}}{\left(2n\right)!}=\frac{e^{\omega x}+e^{-\omega x}}{2}=\frac{e^{-x}+e^{x}}{2}=\color{red}{\cosh\left(x\right)}.$$
