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.
 A: Hint:
$$\cosh x=\sum _{n=0}^{\infty }{\frac {x^{2n}}{(2n)!}}=1+\sum _{n=1}^{\infty }{\frac {x^{2n}}{(2n)!}} $$
A: hint: Consider $f(x) = \dfrac{e^x+ e^{-x}}{2}$
A: 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 $\cosh(x)-1$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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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}
A: 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 $$\sum_{n\geq0}\frac{x^{2n}}{\left(2n\right)!}=\frac{e^{\omega x}+e^{-\omega x}}{2}=\frac{e^{-x}+e^{x}}{2}=\color{red}{\cosh\left(x\right)}.$$
