Integral Test

= Simplified Version =

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:

$$\Delta_n = \sum_{k=1}^n f \left({k}\right) - \int_1^n f \left({x}\right) dx$$

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 $$m \left({b - a}\right) \le \int_a^b f \left({x}\right) dx \le M \left({b - a}\right)$$ 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 $$\forall k \in \mathbb{N}^*: f \left({k+1}\right) \le \int_k^{k+1} f \left({x}\right) dx \le f \left({k}\right)$$, as $$\left({k+1}\right) - k = 1$$.

Thus:

$$ $$ $$ $$

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

Also:

$$ $$ $$ $$

Hence the result.