A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum

Let $(a_k)$ be a sequence of reals with $a_k \ge 0$ for every $k \in \mathbb{N}$, let $s_n = \sum_{k<n} a_k$ be its partial sums and let $S = \{\, s_n : n \in \mathbb{N} \,\}$ be the range of $(s_n)$ (Series, partial sums, convergence and the sum, divergence, and the tail series). Then:

1. $(s_n)$ is nondecreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) and $s_n \ge 0$ for every $n$;
2. $\sum a_k$ converges if and only if $S$ is bounded above (Lower bound, bounded below, bounded set), and in that case $\sum_{k=0}^{\infty} a_k \;=\; \sup S ,$ so in particular $s_n \le \sum_{k=0}^{\infty} a_k$ for every $n$;
3. if $S$ is not bounded above then $s_n \to +\infty$ (Divergence to $+\infty$ and to $-\infty$) and $\sum a_k$ diverges.

This is the theorem that makes the nonnegative theory work: for terms of one sign, convergence is a boundedness question and no candidate limit is ever needed. Every comparison test on this page is an application of it.

• "Bounded" and "bounded above" coincide here, and only here. The equivalence used in step 2.2 rests on $s_n \ge 0$, which rests on every term being nonnegative. For a series with terms of both signs the partial sums can be bounded above and still fail to converge, so nothing in this theorem survives the loss of the sign hypothesis. That failure is exhibited by Two series with $a_k \le b_k$ for all $k$, $\sum b_k$ convergent and $\sum a_k$ divergent, when the terms may be negative ↗ on the companion page.

• Claim 3 is a strictly stronger statement than "diverges". Divergence alone permits oscillation (Series, partial sums, convergence and the sum, divergence, and the tail series); for nonnegative terms it cannot occur, and the partial sums necessarily run to $+\infty$. This is what licenses the phrase "the series diverges to $+\infty$" for nonnegative terms, and it is what the Abel-Dini theorem later on this page uses.

• This criterion is the monotone convergence property, worn differently. The proof above is monotone convergence for $\mathbb{R}$ applied to the nondecreasing sequence of partial sums, and nothing is lost going back the other way. Given a nondecreasing sequence $(x_n)$ of reals, put $y_n := x_n - x_0 \ge 0$ and let $a_0 := 0$, $a_{k+1} := y_{k+1} - y_k \ge 0$; then $s_{n+1} = y_n$ (Series, partial sums, convergence and the sum, divergence, and the tail series), the partial sums are bounded exactly when $(x_n)$ is bounded above, and claim 1 returns the convergence of $(y_n)$ and so of $(x_n)$. Testing boundedness of partial sums is therefore not a device special to series. Read in the vocabulary of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness it is the property (MCT), which in an arbitrary ordered field already forces the Archimedean property on its own (The monotone convergence property alone forces the Archimedean property, so it carries no separate Archimedean hypothesis) and with it the least-upper-bound property (The monotone convergence property plus the Archimedean property imply the least-upper-bound property). The translation just given is carried out in $\mathbb{R}$, since Series, partial sums, convergence and the sum, divergence, and the tail series is stated for sequences of reals and this library defines no series over a general ordered field.