Assume $|a_k|\leq b_k$ for all $k>n$ . Then we define $$s_k:=\sum_{i=k}^\infty |a_i|$$ and $$t_k:=\sum_{i=k}^\infty b_i.$$ Obviously $s_k\leq t_k$ for all $k>n$ . Since by assumption $(t_k)$ is convergent $(t_k)$ is bounded and so is $(s_k)$ . Also $(s_k)$ is monotonic and therefore . Therefore $\sum_{i=0}^\infty a_i$ is absolutely convergent.

Now assume $b_k\leq a_k$ for all $k>n$ . If $\sum_{i=k}^\infty b_i$ is divergent then so is $\sum_{i=k}^\infty a_i$ because otherwise we could apply the test we just proved and show that $\sum_{i=0}^\infty b_i$ is convergent, which is is not by assumption.