The integral test: for $f \ge 0$ nonincreasing on $[0,\infty)$, $\sum_k f(k)$ converges if and only if the sequence $\bigl(\int_0^N f\bigr)_N$ is bounded, with $\int_0^N f \le \sum_{k<N} f(k) \le f(0) + \int_0^N f$

Let $f : [0,\infty) \to \mathbb{R}$ be nonincreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$, with the dictionary to monotone sequences) with $f(t) \ge 0$ for every $t \ge 0$. For $N \in \mathbb{N}$ write

$J_N \;:=\; \int_0^{N} f ,$

which is defined for every $N$: for $N \ge 1$ the restriction of $f$ to $[0,N]$ is monotone, hence bounded and integrable (A monotone function on $[a,b]$ is Riemann integrable: for the uniform partition into $N$ parts the upper minus lower sum telescopes to $|f(b) - f(a)|\,(b-a)/\iota(N)$), and $J_0 = \int_0^0 f = 0$ (The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$). Here $N$ inside the integral means the canonical natural $\iota(N) \in \mathbb{R}$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field), as everywhere in this library. Let $s_N := \sum_{k<N} f(k)$ be the partial sums of $\sum_k f(k)$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion), the index $k$ ranging over $\mathbb{N}$, which contains $0$. Then:

1. The bracket. For every $N \in \mathbb{N}$, $J_N \;\le\; s_N \;\le\; f(0) + J_N .$
2. The test. $(J_N)_{N\in\mathbb{N}}$ is nondecreasing, and $\sum_k f(k)$ converges if and only if the set $\{\,J_N : N \in \mathbb{N}\,\}$ is bounded above (Lower bound, bounded below, bounded set).

The conclusion is about a sequence of proper integrals, and that is deliberate. This library has not defined $\int_0^{\infty} f$ at this point in the reading order — improper integrals are developed on a later page — so the statement that a reader may expect, "$\sum_k f(k)$ converges if and only if $\int_0^{\infty} f$ converges", is not available and is not made. What is proved is the statement above, which is what that one abbreviates; the later page is where the two are identified.

The index starts at $0$. Both the sum and the integral begin at $0$, because $\mathbb{N}$ contains $0$ and a sequence is a function on $\mathbb{N}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences). The classical statement, which starts at $1$, is the statement about the first tail of $\sum_k f(k)$ and is not the statement above.

• Why the bracket is stated with $f(0)$ and not with a tail. The two sums in step 3.1 differ by exactly the first term $f(0)$ and the last term $f(N)$; the clean two-sided statement that survives at every $N \ge 0$, including $N = 0$ where it reads $0 \le 0 \le f(0)$, is the one displayed in claim 1.

• The version beginning at $k = 1$ is a statement about a tail. For the family $(f(k))_{k \ge 1}$ the same argument on $[1,N]$ gives $\int_1^N f \le \sum_{k=1}^{N-1} f(k) \le f(1) + \int_1^N f$, and convergence of the two series is equivalent by A series converges iff each of its tail series converges, and the sum splits as $s_N$ plus the $N$-th tail. Nothing above silently starts at $1$, and a reader comparing with a classical text should check which convention that text uses for $\mathbb{N}$.

• Monotonicity of $f$ is used exactly twice, in step 1.1, to bracket $f$ on a unit interval by its two endpoint values, and through A monotone function on $[a,b]$ is Riemann integrable: for the uniform partition into $N$ parts the upper minus lower sum telescopes to $|f(b) - f(a)|\,(b-a)/\iota(N)$ to know that $f$ is integrable on $[0,N]$ at all. Nonnegativity is used in three named places: to make $(J_N)$ nondecreasing in step 3.2, to pass from $s_{N+1}$ to $s_N$ in step 4.1, and to apply A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, whose own hypothesis is that the terms are nonnegative. No claim is made here about what happens to the test if that hypothesis is dropped.