Signs of Maclaurin coefficients of $\frac{1}{e^x+q}$ My question is based on question by Robert Israel. There is only one answer by Lucia and also two useful comments (by Lucia too), from which we understand, that if
$$\frac{1}{e^x+q}=\sum\limits_{n=0}^{\infty}F_{n}\frac{x^n}{n!}$$
so signs of $F_{n}$ be the same as signs of
$$\cos\left((n+1)\arctan\left(\frac{\pi}{\log q}\right)\right)$$
for $q=3$.
Is it obvious from Lucia's answer (not comments)? If not, why is that true? Is it true for any $q$?
 A: To determine the asymptotic behaviour of the MacLaurin coefficients $F_n$ of a function $f(z)$, the answerer implicitly uses an approach due to Darboux (see Olver, Asymptotic and Special Functions (1997) 4.9.2, p. 310). Suppose that $r$ is the distance from the origin of the nearest singularity of $f(z)$, if we can find a ``comparison function'' $g(z)$ with the properties


*

*$g(z)$ is isomorphic in $0<\left|z\right|<r$

*$f(z)-g(z)$ is continuous  in $0<\left|z\right|\le r$

*The coefficients $b_n$ in the Laurent expansion
\begin{equation}
  g(z)=\sum_{-\infty}^\infty b_nz^n
 \end{equation} 
have known asymptotic behaviour,
then, for $n\to\infty$,
\begin{equation}
 F_n=b_n+o\left( r^{-n} \right)
\end{equation} 
The function \begin{equation}
 f(z)=\frac{1}{e^{z}+q}
 \end{equation} 
is meromorphic and has poles at 
\begin{equation}
 z_n= \ln q+\left( 2n+1 \right)i\pi 
\end{equation} 
where $n$ is an integer. The nearest poles are $z_1=\ln q+i\pi $ and $z_{-1}=\bar{z_1}$. Corresponding residues are both $-1/q$. We choose 
\begin{equation}
 g(z)=-\frac{1}{q}\left( \frac{1}{z-z_1}+\frac{1}{z-z_{-1}} \right)
\end{equation} 
it satisfies the required conditions with
\begin{equation}
 g(z)=\frac{1}{q}\sum_{n=0}^\infty  \left( z_1^{-n-1}+z_{-1}^{-n-1}\right)z^n
\end{equation} 
and thus, by denoting $z_1=\rho e^{i\varphi}$,
\begin{equation}
 b_n=\frac{2}{q}\rho^{-n-1}\cos\left( n+1 \right)\varphi
\end{equation} 
then 
\begin{equation}
 F_n\sim\frac{2}{q}\rho^{-n-1}\cos\left( n+1 \right)\varphi
\end{equation}
Their signs are thus approximatively given by that of $\cos\left(\left(  n+1 \right)\arctan\frac{\pi}{\ln q} \right)$
Finally, it is remarked that, when $q=3$, we have $\varphi\simeq \frac{11\pi}{28}$ with a good accuracy. This explains the near 28-periodicity of the coefficients.
