Calculate sum of infinite series by solving a differential equation Calculate the sum of the infinite series 
$$\sum_{n=0}^{\infty}\frac{1}{(3n)!}$$ by solving an aptly chosen differential equation.
I know that one can solve a differential equation by assuming that we can write the solution as a power series in the form
$$y(x)=\sum_{n=0}^\infty a_n(x-x_0)^n$$ and then find all the different values for $a_n$'s. I'm trying to figure out how to to it the other way around? How am I supposed to find the differential equation when I have the infinite sum already?
Update
I've started off supposing there exists some solution, to the differential equation of the form
$$p(x)y''(x)+q(x)y'(x)+r(x)y(x)=0,$$ that can be written as 
$$y(x)=\sum_{n=0}^\infty a_n(x-x_0)^n. $$
Since I already know this solution should be of the form $$y(x)=\sum_{n=0}^\infty \frac{1}{(3n)!}$$ I know that $a_n=\frac{1}{(3n)!}, x=1, x_0=0$. Furthermore, writing $$y'(x)=\sum_{n=1}^\infty n\ a_n(x-x_0)^{n-1} \ \ and \ \ y''(x)=\sum_{n=2}^\infty n(n-1)\ a_n(x-x_0)^{n-2}. $$ and filling this in the differential equation I finally end up with the equation
$$\sum_{n=0}^\infty \left(p(x)\frac{(n+2)(n+1)}{(3(n+2))!} + q(x)\frac{(n+1)}{(3(n+1))!}+r(x)\frac{1}{(3n)!} \right)=0$$
I think that all the functions $p(x), q(x) \ $and $ r(x)$ should be evaluated at $x=1$. However, I don't know how to proceed. 

From a polynomial or power series $f(x)$ you can "punch out" the odd degree terms by taking the even part $\frac12(f(x)+f(-x))$. In a similar fashion you can produce "holes" with period $3$ in the coefficient sequence by combining $f(e^{ik\frac{2\pi}3}x)$, $k=-1,0,1$.
As the remaining coefficients in the given series can be interpreted to be every third coefficient of the exponential series,
it can be written the evaluation at $x=1$ of $$
\frac13(e^x+e^{wx}+e^{\bar wx})
$$
with $w^3=1$, i.e., $w=e^{i\frac{2\pi}3}=\frac12(-1+i\sqrt3)$.
This can now easily be transformed into a linear ODE of order 3. (Which, arguably, is backwards from what is intended for the task. So you got to find an argument directly starting from the periodicity of $n!a_n$.)
