= Solution
Index the terms from $n=0$:
$$
S(x)=\sum_{n=0}^{\infty}\frac{x^{n+4}}{(n+3)n!}.
$$
Its absolute successive-term ratio tends to zero for every fixed $x$, so it converges everywhere. Divide by $x$ temporarily and use the <exponential series divided by a shifted index> with $k=3$:
$$
\frac{S(x)}x=\int_0^x t^2e^t\,dt
=\big[e^t(t^2-2t+2)\big]_0^x.
$$
Therefore
$$
\boxed{S(x)=x\big[(x^2-2x+2)e^x-2\big].}
$$
At $x=0$ both the original series and this formula are zero, so the temporary division loses no value. The termwise <integration> is justified by uniform convergence of the exponential series on the finite <integration> interval.
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