<s>[INST] 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.
 [/INST] $$f(x)= \sum_{n=3}^{\infty} \frac{x^{n}}{(n+1) ! n 3^{n-2}}$$
$$f^{'}(x)= \sum_{n=3}^{\infty} \frac{nx^{n-1}}{(n+1) ! n 3^{n-2}}= \frac{3^3}{x^2}\sum_{n=3}^{\infty} \frac{x^{n+1}}{(n+1) !  3^{n+1}} = \frac{3^3}{x^2} \left( e^{\frac{x}{3}}-1- \frac{x}{3}\right)$$
</s>