After trying some more questions on Series I'm coming across problems that are rather similar but can't quite grasp what the question is asking for.
The question is as follows:
Write the first four terms of the series expansion of $e^x$ and hence find $$\sum_{r=1}^{\infty}\frac{1}{(r+1)!3^r}$$
The series expansion of $e^x$ is: $$e^x = 1 + x + \frac{x^2}{!2} + \frac{x^3}{!3} + \frac{x^4}{!4}$$
However I can't seem seem to integrate that with the question in any way and am kindly asking for help.
Note that our expression is equal to $$3\sum_1^{\infty}\frac{1}{(r+1)!3^{r+1}},$$ and $$\sum_{1}^\infty \frac{1}{(r+1)!3^{r+1}}=\sum_{n=0}^\infty \frac{x^n}{n!}-1-\frac{x}{1!},$$ where $x=1/3$.