Topic 10.4 Integral Test for Convergence: use the convergence of a related improper integral to decide convergence of a series with positive, decreasing terms (BC).

What this topic is asking

The College Board (Topic 10.4, BC only) gives the integral test, which links a series to an improper integral (Topic 6.13). When the terms come from a positive, decreasing function, the series and the integral of that function converge or diverge together, so you can decide the series by evaluating an integral.

The test and its hypotheses

Why the integral and series agree

The picture behind the test is comparing the series to areas under the curve $y = f(x)$. Because $f$ is decreasing, you can box the series between left- and right-endpoint rectangle sums of the integral. The rectangles for $\sum a_n$ are trapped above and below the area $\int f\,dx$, so if the area is finite (integral converges), the boxed sum is finite too, and if the area is infinite, the sum is forced to be infinite as well. The positive-and-decreasing hypotheses are what make the rectangles line up cleanly with the curve; without them the comparison breaks. This is also why the integral's exact value is not the series sum: the rectangles only bound the area, they do not equal it.

A worked application

Checking "decreasing" properly

A free-response integral-test question expects you to justify the hypotheses, especially that $f$ is decreasing. The clean way is to show $f'(x) < 0$ for $x$ beyond some point, or to argue from the structure of $f$ (for example, $\frac{1}{x^p}$ with $p > 0$ is clearly decreasing). The test only needs $f$ to be eventually decreasing, so you may start the integral at a larger $N$ if the early terms misbehave, since finitely many terms never affect convergence. Stating "positive, continuous, and decreasing on $[N, \infty)$" before evaluating the integral earns the hypothesis points and is required for a complete answer.

What the integral test proves about p-series

The integral test's most important payoff is the $p$-series rule. Applying it to $f(x) = \frac{1}{x^p}$ gives $\int_1^\infty x^{-p}\,dx$, which converges exactly when $p > 1$ and diverges when $p \le 1$, by the improper-integral benchmark of Topic 6.13. Therefore $\sum\frac{1}{n^p}$ converges if and only if $p > 1$, the result you use constantly as a comparison benchmark (Topic 10.5). The harmonic series $\sum\frac{1}{n}$ is the $p = 1$ borderline case, which diverges, matching $\int_1^\infty\frac{1}{x}\,dx = \infty$. So the integral test both decides individual series and establishes the standard family that all the comparison tests lean on.