$c(n)$ is trend, $r(n)$ is fluctuation. Should $\text{cov}[c(n),r(n)]/\text{var}[r(n)]$ be close to zero? Suppose $y(n)$ is a random time series given as function of the discrete-time variable $n$. Suppose we can decompose it into $y(n) = c(n) + r(n)$, where $r(n)$ is a strict stationary residual representing fluctuations and $c(n)$ is a component exhibiting a slowly-varying first-order nonstationary behavior (i.e. a trend).
Given what was defined, can we say that $\text{cov}[c(n),r(n)]/\text{var}[r(n)]$ (*) should be approximately zero? Why? (**) If no, what can we expect from that ratio?
(*) = $\text{cov}[\cdot]$ and $\text{var}[\cdot]$ stand for the stochastic covariance and variance operators
(**) = I'm asking such a question because, in the application I've been working on, in nearly all cases such a fraction is close to zero.
 A: I don't think your conclusion is true in the generality with which the question is posed.  Here is a class of simple counterexamples.
Let $X$ be a Bernoulli$(1/2)$ variable.  Pick a small number $\beta$ (to determine how slowly $c$ will vary) and any number $\sigma$ intended to become large.  Define $$c(n)=\beta n+\sigma X$$ and $$r(2n)=X-1/2,$$ $$r(2n+1)=1/2-X$$ for all integers $n$.  Since for any realization the first differences of $c$ all equal $\beta$ and the expected first differences are $\beta$, $c$ can be made as slowly varying as you wish by picking sufficiently small $\beta$. Because $X-1/2$ and $1/2-X$ are identically distributed, $r$ is strictly stationary and certainly "fluctuates," since it continually alternates signs and is symmetric about zero.  But
$$\operatorname{Cov}(c(n), r(n)) = \pm \sigma \operatorname{Var}(X) = \pm \sigma/4$$
shows this covariance can be arbitrarily large relative to the variance of $r(n)$, which is just $1/4$.

The red and blue plots depict the only two possible realizations of the process $y=c+r$.  Each has probability $1/2$.  Note the different scales on the axes: there is a slow linear trend $c$ surrounded by regular oscillations $r$.  Fix any index $n$: when $c(n)$ goes up, $r(n)$ reverses, creating a strong correlation between $c(n)$ and $r(n)$.
A: The trend $c(n)$ does not have to be a linear trend, or monotonic. The trend, in this case, can be seen as a slowly-varying component in comparison to $r(n)$ that carries information about global changes in the time series $y(n)$. The time series with trend and fluctuation superimposed would like this:

Where the trend $c(n)$ is a part of a sine function, and the fluctuation $r(n)$ is zero-mean white Gaussian noise (See Matlab code below). As you can see in the matlab script below, if I run 1000 realizations of $y(n)$ and estimate stat = cov(c,r)/var(r), the following histogram of "stat" is obtained:

We have evidences the expected value of cov(c,r)/var(r) is zero, and it does not fluctuates to far from zero either. Why? Perhaps due to the fact $c(n)$ can be considered to be approximately constant in comparison to $r(n)$ in short intervals, and due to this property of covariance $cov(\sum_i X_i, \sum_j Y_j) = \sum_i \sum_j cov(X_i , Y_j )$ it should be, in this case, approximately zero? (as the covariance with a constant is zero).
number_of_samples = 1000;  % number of time instants to consider
initial_time = 1;
final_time = 100;
n = linspace(initial_time,final_time,number_of_samples); 
% Amplitude of r(n) and c(n) set to be the same
amplitude_term = 3; 
% Run 1000 realizations of the process
for i = 1:1000 
r = amplitude_term*normrnd(0,1,1,length(n)); 
% trend c(n) doesn't have to be monotonic
c = amplitude_term*sin((0.0085)*2*pi*n + rand(1)); 
% time series: superposition of r and c
y = c + r; 
% estimating cov(c,y)/var(r)
stat_to_estimate(i) = mean((c - mean(c)).*(r - mean(r)))/var(r); 
end
hist(stat_to_estimate)
