Cauchy Integral Test
Theorem
Let $f$ be a real function which is continuous, positive and decreasing on the interval $\hointr 1 {+\infty}$.
Let the sequence $\sequence {\Delta_n}$ be defined as:
- $\ds \Delta_n = \sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x$
Then $\sequence {\Delta_n} $ is decreasing and bounded below by zero.
Hence it converges.
Proof
From Darboux's Theorem, we have that:
- $\ds m \paren {b - a} \le \int_a^b \map f x \rd x \le M \paren {b - a}$
where:
- $M$ is the maximum
and:
- $m$ is the minimum
of $\map f x$ on $\closedint a b$.
Since $f$ decreases, $M = \map f a$ and $m = \map f b$.
Thus it follows that:
- $\ds \forall k \in \N_{>0}: \map f {k + 1} \le \int_k^{k + 1} \map f x \rd x \le \map f k$
as $\paren {k + 1} - k = 1$.
Thus:
| \(\ds \Delta_{n + 1} - \Delta_n\) | \(=\) | \(\ds \paren {\sum_{k \mathop = 1}^{n + 1} \map f k - \int_1^{n + 1} \map f x \rd x} - \paren {\sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map f {n + 1} - \int_n^{n + 1} \map f x \rd x\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \map f {n + 1} - \map f {n + 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
Thus $\sequence {\Delta_n}$ is decreasing.
$\Box$
Also:
| \(\ds \Delta_n\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \map f k - \sum_{k \mathop = 1}^{n - 1} \int_k^{k + 1} \map f x \rd x\) | ||||||||||||
| \(\ds \) | \(\ge\) | \(\ds \sum_{k \mathop = 1}^n \map f k - \sum_{k \mathop = 1}^{n - 1} \map f k\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map f n\) | ||||||||||||
| \(\ds \) | \(\ge\) | \(\ds 0\) |
$\Box$
Hence the result.
$\blacksquare$
Also known as
The is also known as the Euler-Maclaurin summation formula, but that result properly refer to a more precise theorem of which this is a simple corollary.
Some sources just use the name integral test.
Motivation
It follows from the that if $f$ is continuous, positive and decreasing on $\hointr 1 \infty$, then the series $\ds \sum_{k \mathop = 1}^\infty \map f k$ and the improper integral $\ds \int_1^{\mathop \to +\infty} \map f x \rd x$ either both converge or both diverge.
So this theorem provides a test for the convergence of both a series and an improper integral.
Source of Name
This entry was named for Augustin Louis Cauchy.
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 13.32$
- 1992: Larry C. Andrews: Special Functions of Mathematics for Engineers (2nd ed.) ... (previous) ... (next): $\S 1.2.2$: Summary of convergence tests: Theorem $1.2$
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Cauchy integral test
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Cauchy integral test