Proof of Integral Test

Statement of integral test

Suppose $f:[1,\infty) \to \mathbb{R}$ is positive and decreasing on $[1,\infty)$. Then $\sum_{k=1}^{\infty} f(k)$ converges if and only if $\int_1^{\infty} f$ converges.

Proof

Let $n\in\mathbb{Z}^+, n\geq 2$. By properties of the Riemann Integral, $\int_1^{n} f$ exists. Partition the interval $[1,n]$ by $\mathcal{P} = \{1<2<3<\cdots <n\}$. We now have

$\sum_{k=2}^{n} \inf_{t\in[k-1,k]} f(t)\cdot[k-(k-1)] \leq \int_1^n f \leq \sum_{k=2}^{n} \sup_{t\in[k-1,k]} f(t)\cdot[k-(k-1)].$

As $f$ is positive and decreasing, it takes it's supremum at the left most point of each interval $t=k-1$ and it's infimum at it's right most point of each interval, $t=k$ for each $k\in\mathbb{Z}^+$. We now have

$\sum_{k=2}^{n} f(k) \leq \int_1^n f \leq \sum_{k=2}^{n} f(k-1).$

This is equivalent to the statement (by adjustment of index)

$\sum_{k=2}^{n} f(k) \leq \int_1^n f \leq \sum_{k=1}^{n-1} f(k).$

Now define for $n \in \mathbb{Z}^+$

$s_n = \sum_{k=1}^{n} f(k)$

and note that

$s_{n+1} - s_n = f(n+1) \geq 0$, and we deduce that $s_n$ is monotone increasing. Now suppose that $\int_1^{\infty} f$ is finite. Then we have by the left hand side of the inequality derived above

$s_n < f(1) + \int_1^{\infty} f$

and this means that $s_n$ is bounded above and monotone increasing, so the infinite sum $\sum_{k=1}^{\infty} f(k)$ converges. Now suppose that $\int_1^{\infty} f$ diverges. Then by the right hand side of the inequality we have

$\int_1^{n} f \leq \sum_{k=1}^{n-1} f(k)$

and this implies that $\sum_{k=1}^{\infty} f(k)$ diverges. This completes the proof.