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.

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$$\bbox[10px,#ffe,border:1px dotted navy]{\ds{%
\mbox{Besides your 'differential equation technique', you can perform a direct evaluation as follows:}}}
$$
\begin{align}
\sum_{n = 1}^{\infty}{x^{2n} \over \pars{2n}!} & =
\sum_{n = 2}^{\infty}{x^{n} \over n!}\,{1 + \pars{-1}^{n} \over 2} =
{1 \over 2}\pars{\sum_{n = 0}^{\infty}{x^{n} \over n!} - 1 - x} +
{1 \over 2}\pars{\sum_{n = 0}^{\infty}{\pars{-x}^{n} \over n!} - 1 + x}
\\[5mm] & =
-1 + {1 \over 2}\,\expo{x} + {1 \over 2}\,\expo{-x} =\
\bbox[#ffe,10px,border:1px dotted navy]{\ds{\cosh\pars{x} - 1}}
\end{align}
