# 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 Upper and Lower Bounds of Integral, 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 integral test 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.