I seem to be stuck on this question. Or rather, not sure if I solved it correctly.
So I know that
$$\frac{1}{(n+1)!}=\frac{1}{(n+1)}*\frac{1}{(n)!}$$
Therefore, I can change $\frac{1}{(n+1)}$ to $1-n+n^2-n^3....$
which would make:
$$f(n)=\frac{(1-n+n^2-n^3...)}{(n)!}$$
Would that be the final answer or is there another way to write $\frac{1}{(n)!}$ to answer this question correctly?
Thanks in advance!
If you want to model $\frac{1}{n+1}$ as a geometric series, then, $$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots $$ so letting $x = -n$ you indeed have $$ \frac{1}{1+n} = 1 - n + n^2 - n^3 \pm \ldots $$ but I don't see how this can help you.
You are looking for a generating function for the sequence $a_n = \frac1{(n+1)!}$, so you get $$ \begin{split} f(x) &= \sum_{k=0}^\infty a_k x^k = \sum_{k=0}^\infty \frac{x^k}{(k+1)!} \\ &= \frac{1}{x} \sum_{k=0}^\infty \frac{x^{k+1}}{(k+1)!} \quad \text{use } m = k+1\\ &= \frac{1}{x} \sum_{m=1}^\infty \frac{x^m}{m!} \\ &= \frac{1}{x} \left[\left(\sum_{m=0}^\infty \frac{x^m}{m!}\right) -1\right] \end{split} $$ Can you finish this by recognizing the sum as a Taylor series for a well-known function?