calculate $\sum_{n=0}^\infty \frac{3^n}{n!(n+3)}$ using power series let $f(x)=\frac{e^x-1-x-\frac{x^2}{2}}{x}$, because $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$, $f$ can be expressed as $$f(x) = \frac{\sum_{n=0}^\infty \frac{x^n}{n!}-1-x-\frac{x^2}{2}}{x}=\frac{\sum_{n=3}^\infty \frac{x^n}{n!}}{x}=\sum_{n=0}^\infty \frac{x^{n+2}}{(n+3)!}$$
the power series converge in $(-\infty, \infty)$ because $\lim_{n\to\infty} \sqrt[n]{\frac{1}{(n+3)!}}=0$ and let $f_n(x) = \frac{x^{n+2}}{(n+3)!} \Longrightarrow f'_n(x) = \frac{x^{n+1}}{(n+1)!(n+3)}$, $\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)!(n+3)}= \sum_{n=0}^\infty f'_n(x)$ also converge in $(-\infty, \infty)$ (for the same reason), hence $$f'(x) = \sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)!(n+3)}$$ by repeating this process once more I get $$f''(x) = \sum_{n=0}^\infty \frac{x^n}{n!(n+3)}$$ and if $x=3$ I get $$\sum_{n=0}^\infty \frac{3^n}{n!(n+3)} = f''(3)$$ which is what was looking for. my problem is that $f$ isn't defined for $x=0$ yet the series does converge for it as $\sum_{n=0}^\infty \frac{0^n}{n!(n+3)}=0$, so was the function $f$ I used wrong? or could it be that I can't differentiate $f$ the way I did?

Try using the following instead so you avoid that mess:
$$x^2e^x=\sum_{n=0}^{\infty} \frac{x^{n+2}}{n!}$$
$$\int x^2e^x \; dx=x^2e^x-2xe^x+2e^x+C=\sum_{n=0}^{\infty} \frac{x^{n+3}}{n!(n+3)}$$
 At $x=3$:
$$9e^3-6e^3+2e^3-2=5e^3-2=\sum_{n=0}^{\infty} \frac{3^{n+3}}{n!(n+3)}$$
Notice that the series on the right is what you're looking for but multiplied by $3^3$, so divide both sides by $27$.
I understand that this approach is slightly different than yours, but I believe this approach is faster and easier to understand so I thought you might appreciate it.  I'm sure you can refer to other answers posted here regarding your confusion with $f(0)$.
$$\boxed{\frac{5e^3-2}{27}}$$
