[proofplan] We prove both parts of the Comparison Test using monotonicity of partial sums and the [monotone convergence theorem for sequences](/theorems/743). For part (i), the partial sums of the smaller series are bounded above by the convergent larger series, so they converge. Part (ii) is the contrapositive: the partial sums of the larger series are bounded below by those of the divergent smaller series, forcing divergence. [/proofplan] [step:Prove that convergence of the larger series implies convergence of the smaller series] Let $S_N = \sum_{j=1}^{N} a_j$ and $T_N = \sum_{j=1}^{N} b_j$. Assume $\sum_{j=1}^{\infty} b_j$ converges to $B \in \mathbb{R}$. Since $a_j \geq 0$, the sequence $(S_N)$ is non-decreasing. For $N \geq J$, split the partial sum and use $a_j \leq b_j$ for $j \geq J$: \begin{align*} S_N = \sum_{j=1}^{J-1} a_j + \sum_{j=J}^{N} a_j \leq \sum_{j=1}^{J-1} a_j + \sum_{j=J}^{N} b_j \leq \sum_{j=1}^{J-1} a_j + B. \end{align*} The sequence $(S_N)$ is non-decreasing and bounded above by $\sum_{j=1}^{J-1} a_j + B$. By the [monotone convergence theorem for sequences](/theorems/743), $(S_N)$ converges, so $\sum_{j=1}^{\infty} a_j$ converges. [guided] The strategy is straightforward: a non-decreasing sequence that is bounded above must converge. The partial sums $S_N = \sum_{j=1}^{N} a_j$ are non-decreasing because each $a_j \geq 0$, so $S_{N+1} = S_N + a_{N+1} \geq S_N$. To find an upper bound, we exploit $a_j \leq b_j$ for $j \geq J$. For $N \geq J$: \begin{align*} S_N = \sum_{j=1}^{J-1} a_j + \sum_{j=J}^{N} a_j \leq \sum_{j=1}^{J-1} a_j + \sum_{j=J}^{N} b_j \leq \sum_{j=1}^{J-1} a_j + B. \end{align*} The first $J-1$ terms contribute a fixed finite constant, and the tail is dominated by the convergent series $\sum b_j$ with sum $B$. By the [monotone convergence theorem for sequences](/theorems/743), $\sum_{j=1}^{\infty} a_j$ converges. [/guided] [/step] [step:Prove that divergence of the smaller series implies divergence of the larger series] Assume $\sum_{j=1}^{\infty} a_j$ diverges. Since $a_j \geq 0$, the partial sums $S_N \to \infty$. For $N \geq J$, the inequality $b_j \geq a_j$ for $j \geq J$ gives \begin{align*} T_N = \sum_{j=1}^{J-1} b_j + \sum_{j=J}^{N} b_j \geq \sum_{j=1}^{J-1} b_j + \sum_{j=J}^{N} a_j = \sum_{j=1}^{J-1} b_j + (S_N - S_{J-1}). \end{align*} As $N \to \infty$, $S_N \to \infty$, so $T_N \to \infty$. Therefore $\sum_{j=1}^{\infty} b_j$ diverges. [/step]