[proofplan] The proof uses the enclosure property for consecutive partial sums of a decreasing [alternating series](/page/Alternating%20Series): the sum $s$ lies between $s_N$ and $s_{N+1}$. Since the difference $s_{N+1}-s_N$ is exactly the first omitted term, the remainder $R_N=s-s_N$ points in the same direction as that omitted term. The same enclosure also bounds the distance from $s_N$ to $s$ by the distance from $s_N$ to $s_{N+1}$, which is $b_{N+1}$. [/proofplan] [step:Enclose the sum between two consecutive partial sums] Fix $N \in \mathbb{N}$. The hypotheses say exactly that \begin{align*} \sum_{n=1}^{\infty} (-1)^{n+1}b_n \end{align*} is a decreasing alternating series with sum $s$ and partial sums $(s_M)_{M=1}^{\infty}$. Therefore, the following enclosure result applies: [citetheorem:9710] Hence $s$ lies between $s_N$ and $s_{N+1}$. [/step] [step:Express the remainder as a fraction of the first omitted term] Define the real number \begin{align*} \Delta_N = s_{N+1}-s_N. \end{align*} By the definition of the partial sums, \begin{align*} \Delta_N = (-1)^{N+2}b_{N+1}. \end{align*} Since $s$ lies between $s_N$ and $s_{N+1}$, there exists a number $\theta_N \in [0,1]$ such that \begin{align*} s = s_N + \theta_N \Delta_N. \end{align*} Indeed, if $\Delta_N=0$, then $s_N=s_{N+1}$ and the enclosure gives $s=s_N$, so we may take $\theta_N=0$. If $\Delta_N \neq 0$, then \begin{align*} \theta_N = \frac{s-s_N}{\Delta_N} \end{align*} belongs to $[0,1]$ precisely because $s$ lies between $s_N$ and $s_{N+1}=s_N+\Delta_N$. Thus \begin{align*} R_N = s-s_N = \theta_N\Delta_N = \theta_N(-1)^{N+2}b_{N+1}. \end{align*} [guided] The enclosure from the previous step says that, starting at $s_N$, the point $s$ is somewhere on the closed line segment whose other endpoint is $s_{N+1}$. To encode that statement algebraically, define \begin{align*} \Delta_N = s_{N+1}-s_N. \end{align*} This is the signed displacement from the $N$th partial sum to the next partial sum. By subtracting the two finite sums, every term cancels except the newly added term: \begin{align*} \Delta_N = \sum_{n=1}^{N+1} (-1)^{n+1}b_n - \sum_{n=1}^{N} (-1)^{n+1}b_n = (-1)^{N+2}b_{N+1}. \end{align*} Now use the fact that $s$ lies between $s_N$ and $s_{N+1}$. If $\Delta_N=0$, then $s_N=s_{N+1}$, and the only real number between them is $s_N$ itself. Hence $s=s_N$, so $R_N=0$, and we may write $s=s_N+0\cdot \Delta_N$. If $\Delta_N \neq 0$, define \begin{align*} \theta_N = \frac{s-s_N}{\Delta_N}. \end{align*} Because $s$ lies between $s_N$ and $s_N+\Delta_N$, the signed displacement $s-s_N$ is a nonnegative fraction of the signed displacement $\Delta_N$. Therefore $\theta_N \in [0,1]$, and \begin{align*} s = s_N+\theta_N\Delta_N. \end{align*} Subtracting $s_N$ from both sides gives \begin{align*} R_N = s-s_N = \theta_N\Delta_N. \end{align*} Substituting the value of $\Delta_N$ yields \begin{align*} R_N = \theta_N(-1)^{N+2}b_{N+1}. \end{align*} This formula contains both conclusions: the factor $\theta_N$ controls the size of the remainder, and the factor $(-1)^{N+2}b_{N+1}$ controls its sign. [/guided] [/step] [step:Read off the magnitude and sign of the remainder] Since $\theta_N \in [0,1]$ and $b_{N+1} \ge 0$, the formula \begin{align*} R_N = \theta_N(-1)^{N+2}b_{N+1} \end{align*} gives \begin{align*} |R_N| = \theta_N b_{N+1} \le b_{N+1}. \end{align*} It also gives \begin{align*} (-1)^{N+2}R_N = \theta_N b_{N+1} \ge 0. \end{align*} Therefore, if $R_N \neq 0$, then $R_N$ has the same sign as $(-1)^{N+2}b_{N+1}$. Since $N \in \mathbb{N}$ was arbitrary, both conclusions hold for every $N \in \mathbb{N}$. [/step]