[proofplan] We prove convergence of the alternating series by analyzing the even and odd partial sums separately. The even partial sums form a non-decreasing bounded sequence, and the odd partial sums form a non-increasing bounded sequence; both converge by the [monotone convergence theorem for sequences](/theorems/743). The hypothesis $b_j \to 0$ forces their limits to coincide, establishing convergence. The error bound follows from the same monotonicity structure applied to the tail of the series. [/proofplan] [step:Show that even partial sums are non-decreasing and odd partial sums are non-increasing] Define $S_N = \sum_{j=1}^{N} (-1)^{j-1} b_j$. For the even partial sums: \begin{align*} S_{2n+2} = S_{2n} + (b_{2n+1} - b_{2n+2}). \end{align*} Since $(b_j)$ is decreasing, $b_{2n+1} \geq b_{2n+2}$, so $S_{2n+2} \geq S_{2n}$. Thus $(S_{2n})_{n=1}^{\infty}$ is non-decreasing. For the odd partial sums: \begin{align*} S_{2n+3} = S_{2n+1} - (b_{2n+2} - b_{2n+3}). \end{align*} Since $b_{2n+2} \geq b_{2n+3}$, we have $S_{2n+3} \leq S_{2n+1}$. Thus $(S_{2n+1})_{n=0}^{\infty}$ is non-increasing. [/step] [step:Establish bounds and apply the [monotone convergence theorem for sequences](/theorems/743)] For all $n \geq 1$, the even partial sums satisfy $S_{2n} \geq S_2 = b_1 - b_2 \geq 0$ (since $b_1 \geq b_2$) and \begin{align*} S_{2n} = S_{2n+1} - b_{2n+1} \leq S_{2n+1} \leq S_1 = b_1. \end{align*} So $(S_{2n})$ is non-decreasing and bounded above by $b_1$. Similarly, the odd partial sums satisfy $S_{2n+1} \geq S_{2n} \geq 0$ and $S_{2n+1} \leq S_1 = b_1$, so $(S_{2n+1})$ is non-increasing and bounded below. By the [monotone convergence theorem for sequences](/theorems/743): \begin{align*} S_{2n} \to L \in \mathbb{R}, \qquad S_{2n+1} \to M \in \mathbb{R}. \end{align*} [/step] [step:Use $b_j \to 0$ to show the even and odd limits coincide] The difference between consecutive odd and even partial sums is \begin{align*} S_{2n+1} - S_{2n} = b_{2n+1} \to 0 \quad \text{as } n \to \infty. \end{align*} Taking limits: $M - L = 0$, so $M = L$. Denote this common limit by $S$. Since every partial sum $S_N$ is either an even or odd partial sum, and both subsequences converge to $S$, the full [sequence](/pages/1149) $(S_N)$ converges to $S$. [/step] [step:Derive the error bound $|S - S_N| \leq b_{N+1}$] For even $N = 2n$: the tail $S - S_{2n} = \sum_{j=2n+1}^{\infty} (-1)^{j-1} b_j$ is itself an alternating [series](/pages/1205) with decreasing terms tending to zero. By the same monotonicity argument applied to this tail, $0 \leq S - S_{2n} \leq b_{2n+1}$. For odd $N = 2n+1$: the tail $S_{2n+1} - S = \sum_{j=2n+2}^{\infty} (-1)^{j} b_j = b_{2n+2} - b_{2n+3} + \cdots$ satisfies $0 \leq S_{2n+1} - S \leq b_{2n+2}$. In both cases, $|S - S_N| \leq b_{N+1}$. [guided] Why does the error bound hold? Consider the even case $N = 2n$. The remainder is \begin{align*} S - S_{2n} = b_{2n+1} - b_{2n+2} + b_{2n+3} - b_{2n+4} + \cdots \end{align*} This is an alternating series with decreasing non-negative terms tending to zero. The first term is $b_{2n+1}$, and the same argument that showed the original series has non-negative even partial sums shows $S - S_{2n} \geq 0$. Grouping pairs $(b_{2n+1} - b_{2n+2}) + (b_{2n+3} - b_{2n+4}) + \cdots$ shows positivity, while writing $S - S_{2n} = b_{2n+1} - (b_{2n+2} - b_{2n+3}) - (b_{2n+4} - b_{2n+5}) - \cdots$ shows $S - S_{2n} \leq b_{2n+1}$. The odd case is analogous: $S_{2n+1} - S = b_{2n+2} - b_{2n+3} + \cdots$, giving $0 \leq S_{2n+1} - S \leq b_{2n+2}$. Combining: $|S - S_N| \leq b_{N+1}$ for all $N$. [/guided] [/step]