---
tag: 069HKV4X1E
kind: statement
role: proposition
name: Comparison of a series and an integral
msc: [26A42]
uses: [0QM4TXT3DR]
sources:
  - url: https://en.wikipedia.org/wiki/Integral_test_for_convergence
    locator: Proof
---

## Statement

Let $a$ be a real number, let $n$ be a natural number, and let $f \colon [a, a + n] \to \mathbb{R}$ be a decreasing function.
Then
$$\sum_{i=1}^{n} f(a + i) \le \int_a^{a+n} f(x)\, dx \le \sum_{i=0}^{n-1} f(a + i).$$

## Remarks

On each interval $[a + i, a + i + 1]$ we have $f(a + i + 1) \le f(x) \le f(a + i)$. Integrate and add over $i$.
The integral exists because monotone functions are [[Riemann integrable|04YKAYDQPW]].
With $f(x) = 1/x$ and $a = 1$ it gives $\ln(n + 1) \le H_n \le 1 + \ln n$ for the [[harmonic numbers|0D2Y6G4C4S]].
Mathlib states the first inequality as `AntitoneOn.sum_le_integral`.

## Explanations

### L1 human

The sum of the values at the right ends of unit intervals is an area of rectangles below the graph of a decreasing function.
The sum of the values at the left ends is an area of rectangles above it.