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?
Best Answer
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 $\pi$.