Let $T_n(x)=\sum_{k=0}^{n}\frac{x^k}{k!}$. I'm looking at the function
$$f(x)=\frac{1}{T_n(x)}$$
and I would like to find the power series of this particular function. I know I can use Maclaurin Series to find it, so taking derivatives could help here;
$$f'(x)=\frac{d}{dx}\left[\frac{1}{T_n(x)}\right]=\frac{-T_{n-1}(x)}{T_n(x)^2}$$ Obviously $f(0)=1$ and now $f'(0)=-1$... However I don't want to keep doing this to find a pattern. Is there a quicker way that I'm not seeing?
Edit: It is helpful that $T_{n-k}(0)=1, \forall n>k$ but using the quotient rule for derivatives makes this difficult because how quickly the product rule expands after very few derivatives....
An idea: Using the fact that $1=T_n(x)f(x)$, you can get a recurrence relation on the coefficients $(b_k)_{k}$ of $f\colon x \mapsto \sum_{k=0}^\infty b_k x^k$ by applying a Cauchy product: $$ 1 = \sum_{k=0}^\infty a_k x^k \sum_{k=0}^\infty b_k x^k = \sum_{k=0}^\infty \left(\sum_{\ell=0}^k a_{k-\ell}b_\ell\right)x^k $$ where $a_k = \begin{cases} \frac{1}{k!} & \text{ if } k \leq n\\ 0 & \text{o.w.} \end{cases}$. This gives you $$ 1 = a_{0}b_0 = b_0, \qquad\text{and }\qquad 0 = \sum_{\ell=0}^k a_{k-\ell}b_\ell\quad \forall k \geq 1\ . $$