Impact of sample size on sample covariance From the formula of sample covariance, we see 
$$\operatorname{Cov}(X,Y) = \frac{1}{m} \sum_{i=1}^m (x_i - \bar{x})(y_i  - \bar{y})$$
where $\bar{x}, \bar{y}$ are the sample means. It seems like the sample size $m$ will have an impact on the final outcome. Is the intuition true?
 A: Remember that if you increase the number of data points then you will need actual new data points, so you will have more $x_i$ and $y_i$ values.  This means that the value $m$ in your formula will change, but you will also now have additional terms in your sum, so that will change too.  So yes, the sample covariance will change as you get more values (except in highly unusual cases) because you are using a different data set.
Assuming that the sequences $(X_1,Y_1), (X_2,Y_2), (X_3,Y_3), ...$ are exchangeable, as $m \rightarrow \infty$ the sample covariance will converge to the true covariance between the pairs of values.  So in this case you will generally expect that as $m$ becomes larger the sample covariance will start to become more stable (i.e., change less) and will converge towards a fixed value.

Additional note: It is usual for the sample covariance to be defined with incorporation of Bessel's correction (i.e., using $m-1$ in the denominator instead of $m$).  This is done to ensure that the sample covariance is an unbiased estimator of the true covariance.
