How do I find the sum of a power series $\underset{n=3}{\overset{\infty}{\sum}}\frac{x^n}{(n+1)!n\,3^{n-2}}$? I have found the area of convergence to be $ x \in (-\infty, \infty)$, and this is how far I had gotten before getting stuck:
$$
\begin{aligned}
\sum_{n=3}^{\infty} \frac{x^{n}}{(n+1) ! n 3^{n-2}} &=\sum_{k=0}^{\infty} \frac{x^{k+3}}{(k+4) !(k+3) 3^{k+1}} \\
&=\sum_{k=0}^{\infty} \frac{x^{3} x^{k}}{(k+4) !(k+3) 3 \cdot 3^{k}} \\
&=\frac{x^{3}}{3} \sum_{k=0}^{\infty} \frac{1}{(k+4) !(k+3)}\left(\frac{x}{3}\right)^{k} \\
\text{substituting }u=& \frac{x}{3} \\
&=\frac{x^{3}}{3} \sum_{k=0}^{\infty} \frac{1}{(k+4) !(k+3)} u^{k}
\end{aligned}$$
I do not know how to proceed from here.

Hint
Let $x=3y$ $$A=\underset{n=3}{\overset{\infty}{\sum}}\frac{x^n}{(n+1)!n\,3^{n-2}}=9\sum_{n=3}^\infty \frac{ y^n}{n (n+1)!}=9y\sum_{n=3}^\infty \frac{ y^{n-1}}{n (n+1)!}$$
$$\sum_{n=3}^\infty \frac{ y^{n-1}}{n (n+1)!}=\left(\sum_{n=3}^\infty \frac{ y^{n}}{ (n+1)!}\right)'=\left(\frac 1y\sum_{n=3}^\infty \frac{ y^{n+1}}{ (n+1)!}\right)'=\left(\frac 1y\sum_{n=2}^\infty \frac{ y^{n}}{ n!}\right)'$$
$$\sum_{n=2}^\infty \frac{ y^{n}}{ n!}=-1-y+\sum_{n=0}^\infty \frac{ y^{n}}{ n!}=e^y-y-1$$
