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?
 A: I assume you want the Maclaurin series, i.e. the Taylor series about $0$.
Write $$1 + e^z =  2 (1 + Q(z))$$
where
$$Q(z) = \dfrac{z}{2\cdot 1!} + \dfrac{z^2}{2\cdot 2!} + \dfrac{z^3}{2\cdot 3!} + \ldots $$
So $$ \dfrac{1}{1+e^z} = \dfrac{1}{2(1+Q(z))} = \dfrac{1}{2} \left( 1 - Q(z) + Q(z)^2 - Q(z)^3 + \ldots \right) $$
For $n \ge 1$, the $z^n$ term in the Taylor series comes from the $Q(z)^j$ terms for $j = 1$ to $n$.  Thus the series is $\sum_{n=0}^\infty a_n z^n$ where
$$ \eqalign{
a_0 &= \dfrac{1}{2}\cr
a_1 &= \dfrac{1}{2} \left( -\dfrac{1}{2\cdot 1!}\right) = -\dfrac{1}{4}\cr
a_2 &= \dfrac{1}{2} \left( - \dfrac{1}{2 \cdot 2!} + \dfrac{1}{(2 \cdot 1!)^2} \right) = 0\cr
a_3 &= \dfrac{1}{2} \left( -\dfrac{1}{2 \cdot 3!} + \dfrac{2}{(2\cdot 1!)(2\cdot 2!)} - \dfrac{1}{(2 \cdot 1!)^3}\right) = \dfrac{1}{48}\cr
\ldots}$$
The radius of convergence is the radius of the largest disk around $0$ in which $1/(1+e^z)$ is analytic, i.e. the distance to the closest points where $1 + e^z = 0$.  These points are $\pm i\pi$, so the radius of convergence is $\pi$.
A: 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.
A: To expand on my comment:$$\frac{1}{1+e^z}=\frac{1}{1-(-e^z)}=\sum_{n=0}^\infty (-e^z)^n$$
The series converges if $|-e^z|<1$.
