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:

Thus $\sequence {\Delta_n}$ is decreasing.

Also:

Hence the result.

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.