CONTEXT: Uni question made up by lecturer.
If $b_n \le a_n \le 0$ and $\sum b_n$ is convergent, does this mean $\sum a_n$ is also convergent? I'm pretty sure this is true but can't work out how to apply the comparison test to it.
The comparison test states:
(1) If $\sum_{n=0}^\infty b_n$ is convergent and $0 \le a_n \le b_n$, then $\sum_{n=0}^\infty a_n$ is also convergent.
(2) If $\sum_{n=0}^\infty b_n$ is divergent and $a_n \ge b_n \ge 0$, then $\sum_{n=0}^\infty a_n$ is also divergent.
I can't apply (1) or (2) because $b_n \le a_n \le 0$ does not fit with either of these.
Just apply the comparison test to $-a_n$ and $-b_n$.
Let $\tilde{a_n} = -a_n$ and $\tilde{b_n}=-b_n$.
We have,
$$0\leq \tilde{a_n} \leq \tilde{b_n}$$
Let $B=\sum b_n$ which is finite by assumption.
$$\sum \tilde{b_n} = \sum -b_n = -\sum b_n = -B$$
so $\tilde{b_n}$ is convergent. Apply the comparison test to see that $\tilde{a_n}$ is convergent.