[proofplan] The partial sums of a positive series form an increasing sequence because each new term is nonnegative. One direction follows from the elementary boundedness of convergent real sequences. For the converse, boundedness gives a least upper bound $L$ of the set of partial sums, and monotonicity forces the partial sums to approach $L$. [/proofplan] [step:Show that the partial sums form an increasing sequence] For $N,M \in \mathbb{N}$ with $N < M$, the difference of partial sums is \begin{align*} S_M - S_N = \sum_{n=N+1}^{M} a_n. \end{align*} Since $a_n \geq 0$ for every $n \in \mathbb{N}$, the finite sum on the right is nonnegative. Hence $S_N \leq S_M$ whenever $N < M$, so $(S_N)_{N \in \mathbb{N}}$ is increasing. [/step] [step:Derive boundedness of partial sums from convergence of the series] Assume that the series $\sum_{n=1}^{\infty} a_n$ converges in $\mathbb{R}$. By definition, the sequence of partial sums $(S_N)_{N \in \mathbb{N}}$ converges to some $L \in \mathbb{R}$. Apply convergence with $\varepsilon = 1$. There exists $N_0 \in \mathbb{N}$ such that for every $N \geq N_0$, \begin{align*} |S_N - L| < 1. \end{align*} Therefore $S_N < L+1$ for every $N \geq N_0$. Define \begin{align*} B := \max\bigl(\{L+1\} \cup \{S_N : N \in \mathbb{N},\ 1 \leq N < N_0\}\bigr). \end{align*} The set inside the maximum is finite and nonempty, so $B \in \mathbb{R}$ is well-defined. If $N < N_0$, then $S_N \leq B$ by the definition of $B$. If $N \geq N_0$, then $S_N < L+1 \leq B$. Hence $S_N \leq B$ for every $N \in \mathbb{N}$, so $\{S_N : N \in \mathbb{N}\}$ is bounded above in $\mathbb{R}$. [/step] [step:Use the least upper bound to prove convergence from boundedness] Assume that the set \begin{align*} A := \{S_N : N \in \mathbb{N}\} \end{align*} is bounded above in $\mathbb{R}$. Since $A$ is nonempty and bounded above, the completeness property of $\mathbb{R}$ gives a least upper bound $L := \sup A \in \mathbb{R}$. We prove that $S_N \to L$. Let $\varepsilon > 0$. Since $L$ is the least upper bound of $A$, the number $L-\varepsilon$ is not an upper bound for $A$. Therefore there exists $N_0 \in \mathbb{N}$ such that \begin{align*} S_{N_0} > L-\varepsilon. \end{align*} For every $N \geq N_0$, monotonicity gives $S_N \geq S_{N_0}$, while the definition of $L$ as an upper bound gives $S_N \leq L$. Hence \begin{align*} L-\varepsilon < S_{N_0} \leq S_N \leq L. \end{align*} Thus for every $N \geq N_0$, \begin{align*} 0 \leq L-S_N < \varepsilon, \end{align*} and therefore $|S_N-L| < \varepsilon$. This proves $S_N \to L$ in $\mathbb{R}$. [/step] [step:Conclude the equivalence for the positive series] By definition, the series $\sum_{n=1}^{\infty} a_n$ converges in $\mathbb{R}$ exactly when its sequence of partial sums $(S_N)_{N \in \mathbb{N}}$ converges in $\mathbb{R}$. The previous two steps prove both implications, so the series converges if and only if the set of partial sums is bounded above in $\mathbb{R}$. [/step]