AP Calculus BC

Improper Integral vs Infinite Series

An improper integral extends a definite integral to an unbounded interval or integrand, defined as a limit; a series adds discrete terms. The integral test links them: when the terms come from a positive, continuous, eventually decreasing function, the two share a verdict but almost never a value.

Improper integral

Use when: The interval runs to infinity, or the integrand blows up somewhere inside it, so the value has to come from a limit of ordinary definite integrals.

Infinite series

Use when: The quantity is a sum of terms indexed by whole numbers, so the verdict has to come from a convergence test rather than from an antiderivative.

Side by side

	Improper integral	Infinite series
Written as	$\int_1^{\infty} f(x)\,dx$	$\sum_{n=1}^{\infty} a_n$
Defined by	$\lim_{b \to \infty} \int_1^{b} f(x)\,dx$	$\lim_{N \to \infty} \sum_{n=1}^{N} a_n$
Evaluated with	An antiderivative and one limit	A closed form for the partial sums, which in practice only geometric and telescoping series supply, or recognition of a known Maclaurin series evaluated at a point
Same rule, both objects	$\int_1^{\infty} \frac{dx}{x^2} = 1$	$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.645$
Role in the integral test	Supplies the convergence verdict, nothing more	Inherits the verdict and keeps its own value

Both objects are limits, but of different things. The improper integral $\int_1^{\infty} f(x)\,dx$ means $\lim_{b \to \infty} \int_1^{b} f(x)\,dx$ , a limit of areas over longer and longer intervals. The series $\sum_{n=1}^{\infty} a_n$ means the limit of the partial sums $S_N = a_1 + a_2 + \cdots + a_N$ . One sweeps continuously, the other steps.

f \text{ positive, continuous, decreasing on } [N,\infty), \quad a_n = f(n) \;\Longrightarrow\; \sum_{n=N}^{\infty} a_n \text{ and } \int_N^{\infty} f(x)\,dx \text{ both converge or both diverge}

The verdict is shared; the number is not. For $f(x) = 1/x^2$ the integral comes out as $\int_1^{\infty} \frac{dx}{x^2} = 1$ , while the series comes out as $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.645$ . The reason is geometric: the terms $f(1), f(2), f(3), \dots$ are the heights of left endpoint rectangles of width $1$ , and on a decreasing curve those rectangles stick out above the graph. That traps the sum between the integral and the integral plus the first term, here $1 \le \frac{\pi^2}{6} \le 2$ .

The verdict is not the value

Reporting $\sum_{n=1}^{\infty} \frac{1}{n^2} = 1$ because the matching integral equals $1$ is the error this page exists to prevent. The integral test decides convergence and nothing more. The sums you can actually produce come from geometric series, telescoping series, and known Maclaurin series evaluated at a point.

Frequently asked questions

Can the integral test tell me what the series adds up to?

No. It settles convergence only, and it is not a source of error bounds on the AP exam either. The two remainder bounds you are asked to produce are the alternating series error bound and the Lagrange error bound, and neither one comes from this test. A tail estimate built out of the integral does exist in college calculus, but it is not AP assessed, so treat the integral test as a verdict and nothing else.

Does the function have to be decreasing for every $x$ ?

No, eventually decreasing is enough. Convergence depends only on the tail, so if $f$ decreases on $[N,\infty)$ for some starting value $N$ , apply the test from there. The finitely many terms before $N$ are a finite sum and cannot change the verdict.

Can the integral converge while the series diverges?

Not while all three conditions hold, since the test makes them inseparable. A mismatch means a hypothesis failed: the terms are not positive, the function is not eventually decreasing, or $a_n$ is not $f(n)$ . An alternating series such as $\sum \frac{(-1)^n}{n}$ falls outside the test entirely.

In the CED: Unit 6: Integration and Accumulation, Unit 10: Infinite Sequences and Series (BC)