[proofplan] We compare the sum $s$ with the $N$th partial sum through the remainder $R_N := s-s_N$. The alternating-series remainder estimate gives both the sign of $R_N$ and the bound $|R_N| \le b_{N+1}$. Since the next partial sum satisfies $s_{N+1}=s_N+(-1)^{N+2}b_{N+1}$, the remainder places $s=s_N+R_N$ between the two consecutive partial sums. [/proofplan] [step:Express the next partial sum as the current partial sum plus the next alternating term] Fix $N \in \mathbb{N}$. Define the $N$th remainder $R_N \in \mathbb{R}$ by \begin{align*} R_N := s-s_N. \end{align*} By the definition of the partial sums, \begin{align*} s_{N+1}=s_N+(-1)^{N+2}b_{N+1}. \end{align*} Define the next increment $d_N \in \mathbb{R}$ by \begin{align*} d_N := s_{N+1}-s_N = (-1)^{N+2}b_{N+1}. \end{align*} Thus the desired enclosure is equivalent to proving that $R_N$ lies between $0$ and $d_N$. [/step] [step:Use the alternating remainder estimate to compare the remainder with the next increment] By [citetheorem:9709] applied to the decreasing [alternating series](/page/Alternating%20Series) and its $N$th remainder $R_N$, we have \begin{align*} |R_N| \le b_{N+1}. \end{align*} Moreover $R_N$ has the same sign as $(-1)^{N+2}b_{N+1}=d_N$, unless $R_N=0$. If $d_N \ge 0$, then $d_N=b_{N+1}$ and the sign conclusion gives $R_N \ge 0$. The magnitude bound gives $R_N \le b_{N+1}=d_N$. Hence \begin{align*} 0 \le R_N \le d_N. \end{align*} If $d_N < 0$, then $d_N=-b_{N+1}$ and the sign conclusion gives $R_N \le 0$. The magnitude bound gives $R_N \ge -b_{N+1}=d_N$. Hence \begin{align*} d_N \le R_N \le 0. \end{align*} In both cases, $R_N$ lies between $0$ and $d_N$. [guided] The point of introducing $d_N$ is that the two consecutive partial sums differ by exactly one known term. We have \begin{align*} d_N := s_{N+1}-s_N = (-1)^{N+2}b_{N+1}. \end{align*} So proving that $s$ lies between $s_N$ and $s_{N+1}$ is the same as proving that the displacement from $s_N$ to $s$, namely \begin{align*} R_N := s-s_N, \end{align*} lies between $0$ and the displacement from $s_N$ to $s_{N+1}$, namely $d_N$. We now apply [citetheorem:9709]. Its hypotheses are exactly the hypotheses in force: the series is alternating, the coefficients satisfy $b_n \ge 0$, the sequence $(b_n)$ is decreasing, and $b_n \to 0$. Therefore its conclusion gives \begin{align*} |R_N| \le b_{N+1}. \end{align*} It also gives the sign information: unless $R_N=0$, the sign of $R_N$ is the sign of the first omitted alternating term \begin{align*} (-1)^{N+2}b_{N+1}=d_N. \end{align*} There are two possible orientations. If $d_N \ge 0$, then $d_N=b_{N+1}$. The sign conclusion gives $R_N \ge 0$, while the estimate $|R_N|\le b_{N+1}$ gives $R_N \le b_{N+1}=d_N$. Hence \begin{align*} 0 \le R_N \le d_N. \end{align*} If $d_N < 0$, then $d_N=-b_{N+1}$. The sign conclusion gives $R_N \le 0$, while $|R_N|\le b_{N+1}$ gives $-b_{N+1}\le R_N$. Since $d_N=-b_{N+1}$, this becomes \begin{align*} d_N \le R_N \le 0. \end{align*} Thus, regardless of parity, $R_N$ lies between $0$ and $d_N$. [/guided] [/step] [step:Translate the remainder comparison back to the partial sums] Since $s=s_N+R_N$ and $s_{N+1}=s_N+d_N$, adding $s_N$ to the enclosure of $R_N$ between $0$ and $d_N$ gives that $s$ lies between $s_N$ and $s_{N+1}$. Equivalently, \begin{align*} \min\{s_N,s_{N+1}\} \le s \le \max\{s_N,s_{N+1}\}. \end{align*} Because $N \in \mathbb{N}$ was arbitrary, the conclusion holds for every $N \in \mathbb{N}$. [/step]