In the comparison test, why are the summands in both the smaller and larger series required to be non-negative for all n? Is the algorithm not well-defined if either one of the series has negative summands?  
Thanks,
 A: I'll assume that you're asking about the (direct) comparison test in which given $0\le a_n\le b_n$, the convergence of $\sum_na_n$ is deduced form the convergence of $\sum_nb_n$.
The title seems to imply that the summands of either series could individually be required to be non-negative. However, $a_n\le b_n$ implies that the $b_n$ are non-negative if the $a_n$ are, so there is really only one question here, namely why the $a_n$ are required to be non-negative. Consider $a_n=-1$ and $b_n=1/n^2$. Then $\sum_na_n$ diverges and $\sum_nb_n$ converges despite $a_n\le b_n$.
Now you might say, but why don't we just consider the absolute values then? Consider $a_n=1/n$ and $b_n=(-1)^n/n$. Then $\sum_na_n$ diverges and $\sum_nb_n$ converges despite $\left|a_n\right|\le\left|b_n\right|$.
Now you might say, why don't we consider $\left|a_n\right|\le b_n$ then? We could do that, but it adds nothing new, since we know that every absolutely converent series is convergent, so we might as well apply the standard comparison test to $\left|a_n\right|$ and then deduce the convergence of $a_n$ from its absolute convergence.
