Taylor Series for $\frac{1}{1+e^z}$ and radius of convergence I have done some manipulation and got that $$\frac{1}{1+e^z} = \sum_{n=0}^\infty \frac{n!}{n!+z^n}$$ by the fact that:
$$\frac{1}{1+e^z}= \frac{1}{1+\sum_{n=0}^\infty\frac{z^n}{n!}}=\frac{1}{2}+\frac{1}{1+z}+\frac{1}{1+\frac{z^2}{2}}+\ldots = \frac{1}{2}+\frac{1}{1+z}+\frac{2!}{2!+z^2}+\frac{3!}{3!+z^3}+\ldots$$ $$= \sum_{n=0}^\infty\frac{n!}{n!+z^n}$$
Assuming I did the above right, I am having trouble finding the radius of convergence of the Taylor series given above. I tried the ratio test but got stuck.
Edit: I just realized I did this completely wrong, thinking that I was taking $$\sum\frac{1}{1+\sum_{n=0}^\infty\frac{z^n}{n!}}$$.
Could anyone help me? My question wants me to compute the first four terms of the taylor series for $\frac{1}{1+e^z}$ and find the radius of convergence. Perhaps they do not want me to actually find the explicit form?

One option is: Write
$$\frac{1}{1 + e^z} \cdot (1 + e^z) = 1$$
and expand both of the factors on the l.h.s. in Taylor series" Since $e^z \sim \sum_{k = 0}^{\infty} \frac{1}{k!} z^k$, we have
$$1 + e^z = 1 + \left(1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + O(z^5)\right) = 2 + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + O(z^5).$$
Then, expand the (t.b.d.) Taylor series of $\frac{1}{1 + e^z}$ as $$a_0 + a_1 z + a_2 z^2 + a_3 z^3 + a_4 z^4 + O(z^5).$$
The product of these two series is
$$2 a_0 + (a_0 + 2 a_1) z + \left(\frac{1}{2}a_0 + a_1 + 2 a_2\right) + (\text{third- and fourth-order terms}) + O(z^5),$$
and the polynomial part here must agree with $1$ to fourth order. (Note that we only need consider the involved Taylor expansions here to order $4$ to determine $a_0, \ldots, a_4$, that is, to determine the desired Taylor series to that order.)
So, we get a system of five linear equations in the coefficients $a_0, \ldots, a_4$:
\begin{align}
1 &= 2 a_0 \\
0 &= a_0 + 2 a_1 \\
0 &= \frac{1}{2} a_0 + a_1 + 2 a_2 \\
\vdots &= \vdots .
\end{align}
Solving and substituting gives the desired Taylor series.
