The Integral Test

If $f$ is positive, continuous, and decreasing for $x\geq 1$and $a_{n} = f(n)$, then $\displaystyle\sum_{n=1}^\infty{a_n}$ and $\displaystyle\int_{1}^{\infty}f(x) dx$ either both converge or both diverge.

To prove this theorem we first partition the interval $[1, n]$ into $n-1$ unit intervals. The total areas of the inscribed rectangles and the circumscribed triangles are as follows: $\displaystyle\sum_{i=2}^{n}f(i) = f(2) + f(3) + .... +f(n)$ (inscribed area) $\displaystyle\sum_{i=1}^{n-1}f(i) = f(1) + f(2) + ... + f(n-1)$ (circumscribed area)

The precise area under the graph of $f$ from $x=1$ to $x=n$ lies between the inscribed and circumscribed areas which implies $\displaystyle\sum_{i=2}^{n}f(i) \leq \displaystyle\int_{1}^{n}f(x) dx \leq \displaystyle\sum_{i=1}^{n-1} f(i)$. Using the nth partial sum, $S_n = f(1) + f(2) +... + f(n)$, we can write this inequality as $S_n -f(1) \leq \displaystyle\int_1^n f(x) dx \leq S_{n-1}$. Now assuming that $\int_1^\infty f(x) dx$ converges to L, it follows that for $n \geq 1$, $S_{n} - f(1) \leq L\Rightarrow S_{n} \leq L + f(1)$ . Consequently, ${S_n}$ is bounded and monotonic, and thereby by the Bounded Monotonic Sequence Theorem it converges. So, $\sum a_n$ converges.