$S(x)=\frac{x^4}{2\cdot 4}+\frac{x^6}{2\cdot 4\cdot 6}+\frac{x^8}{2\cdot4\cdot6\cdot8}+\cdots$ Find the sum of $$S(x)=\frac{x^4}{2\cdot 4}+\frac{x^6}{2\cdot 4\cdot 6}+\frac{x^8}{2\cdot4\cdot6\cdot8}+\cdots$$
What I did so far:
It's trivial that
$$S'(x)-xS(x)=\frac{x^3}{2}-\lim_{n\to\infty}\frac{x^{2n}}{n!2^{{n(n+1)}/2}}$$
and $$\lim_{n\to\infty}\frac{x^{2n}}{n!2^{{n(n+1)}/2}}=0$$
Solve this ODE I got
$$S(x)=\frac{x^4}{8}e^x+Ce^x$$
And $S(x)=0$ , So $C=0$.
Finally, I got $S(x)=\frac{x^4}{8}e^x$
The problem is when I expand the function I got into power series, it looks different from the original one. So I may make some mistakes but I can't tell. Please help.

@labbhattacharjee provided a very efficient approach to evaluating the series.  
The approach in the OP did not provide a way forward since the solution to the ODE $S'(x)-xS(x)=\frac12x^3$, $S(0)=0$ is given by
$$\begin{align}
S(x)&=\int_0^x \frac12x'^3 e^{(x^2-x'^2)/2}\,dx' \tag 1\\\\
&=e^{x^2/2}-\frac12x^2-1
\end{align}$$
where we used the integrating factor $\mu(x)=e^{-x^2/2}$ to facilitate generating the solution to the ODE and integration by parts to evaluate the integral in $(1)$.
