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.

Note that your equation should be $f''(x)=f(x)+1$.

Using the series for $e^x$, we get
$$
\begin{align}
e^x&=\sum_{k=0}^\infty\frac{x^k}{k!}\tag{1}\\
e^{-x}&=\sum_{k=0}^\infty(-1)^k\frac{x^k}{k!}\tag{2}
\end{align}
$$
Average $(1)$ and $(2)$
$$
\frac{e^x+e^{-x}}2=\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}\tag{3}
$$
Subtract $1$
$$
\frac{e^x-2+e^{-x}}2=\sum_{k=1}^\infty\frac{x^{2k}}{(2k)!}\tag{4}
$$
$(4)$ can be written as $\cosh(x)-1$.
