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.