Integral Test

Theorem
Let $f$ be a real function which is continuous, positive and decreasing on the interval $\left[{1 \,.\,.\, +\infty}\right)$.

Let the sequence $\left \langle {\Delta_n} \right \rangle$ be defined as:


 * $\displaystyle \Delta_n = \sum_{k \mathop = 1}^n f \left({k}\right) - \int_1^n f \left({x}\right) \rd x$

Then $\left \langle {\Delta_n} \right \rangle$ is decreasing and bounded below by zero.

Hence it converges.

Proof
From Upper and Lower Bounds of Integral, we have that:
 * $\displaystyle m \left({b - a}\right) \le \int_a^b f \left({x}\right) \rd x \le M \left({b - a}\right)$

where:
 * $M$ is the maximum

and:
 * $m$ is the minimum

of $f \left({x}\right)$ on $\left[{a \,.\,.\, b}\right]$.

Since $f$ decreases, $M = f \left({a}\right)$ and $m = f \left({b}\right)$.

Thus it follows that:
 * $\displaystyle \forall k \in \N_{>0}: f \left({k + 1}\right) \le \int_k^{k + 1} f \left({x}\right) \rd x \le f \left({k}\right)$

as $\left({k + 1}\right) - k = 1$.

Thus:

Thus $\left \langle {\Delta_n} \right \rangle$ 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.