Completely monotone condition I've stucked in this problem for a while. So I hope someones could give me suggestions.
Consider the function:
$$
f( r ) = \frac{e^{- br}}{1 + ce^{ - dr}} = \sum\limits_{n = 0}^\infty  {( - c)^n}e^{ - (b + nd)r} 
$$ 
For given $c \in (0,\,1)$ and finite $d>0$, I want to find the condition of $b$ to make $f(.)$ is compelely monotonic function, i.e., 
$$ (-1)^l f^{(l)}(r)= \sum\limits_{n = 0}^\infty  {( - c)^n}(b+nd)^l e^{ - (b + nd)r}  \ge 0$$
for all $l$ and $r$.
For $b = \infty$ I can show that $f(.)$ is completely monotonic. 
Let denote $x_n = ( - c)^n(b+nd)^l e^{ - (b + nd)r}$, we have
$$
x_n - x_{n + 1} = ( - c)^n e^{ - (b + nd)r}(b + nd)^l\left\{ {1 + c{{\left( {1 + \frac{d}{b + nd}} \right)}^l}e^{ - dr}} \right\}
$$
Hence for fixed, positive and finite $b,\, d,\, r$ and $l$, when $n \to \infty$, $(x_n - x_{n+1}) \to 0$. So it seems that $(-1)^l f^{(l)}(r)= \sum\limits_{n = 0}^\infty x_n$ is convergent. Those are all results I got.
 A: For $c\in(0,1)$ and positive $b$ and $d$, the function 
$f$   is never completely monotonic.    
The proof uses a bit of complex function theory,  related to topics in the recent  book Bernstein functions by  Schilling, Song and Vondracek. 
Suppose $f$ is completely monotonic.  By Bernstein's theorem, there is an increasing function $\mu$ on $[0,\infty)$ such that $$ f(r) = \int_0^\infty e^{-tr}\,d\mu(t), \qquad r>0. $$ Then, using Fubini's theorem, the Laplace transform of $f$ has the representation $$ F(s) = \int_0^\infty e^{-sr} f(r)\,dr = \int_0^\infty \int_0^\infty e^{-(s+t)r}\,dr\,d\mu(t) =  \int_0^\infty \frac{d\mu(t)}{s+t}, \quad s>0. $$ This formula defines an analytic extension of $F$ into the upper half plane $\Im s>0$  with the property that $\Im F(s)<0$ there. (Such a function is called a Stieltjes function.)  
However, we can compute $F(s)$ explicitly and conclude otherwise.
 Note that by scaling $r$ we can assume $d=1$:  
$$ F(s) = \int_0^\infty \sum_{k=0}^\infty (-ce^{-r})^k e^{-(s+b)r}\,dr = \sum_{k=0}^\infty \frac{(-c)^k}{s+b+k}. $$ Since $|-c|<1$ this series converges for all complex $s$ not equal to $-b-k$ for some  nonnegative integer $k$. But since $c>0$ it is not a Stieltjes function. E.g., taking  $s=-b-1+i\epsilon$ with $\epsilon>0$, the term for $k=1$  is $ic/\epsilon$. For  small enough $\epsilon>0$ this term dominates and yields $\Im F(-b-1+i\epsilon)>0$.  This contradicts the result from Bernstein's theorem, so $f$ cannot be completely monotonic.  
 (It's clear, however,  you do have complete monotonicity with $c$ of the opposite sign,  i.e., $c\in[-1,0]$. ) 
