Integral Test for Convergence
Introduced| English |
|---|
| Integral Test/ˈɪntɪɡrəl test/ |
A series and an integral that rise or fall together
- If a series' terms come from a nice decreasing function, its fate matches that function's integral.
- The Integral Test 积分判别法 compares $\sum a_n$ with $\int f(x)\,dx$, where $a_n=f(n)$.
- Picture the terms as rectangle areas — they're bounded by the area under the curve.
- So the improper integral converging (or not) forces the series to do the same.
The conditions
- The test needs $f$ to be positive, continuous, and decreasing for $x\ge$ some point.
- Then define $a_n=f(n)$ and compare with the improper integral.
- These conditions let the rectangles sandwich the area under $f$.
- Check all three before applying — a non-decreasing $f$ breaks the comparison.
The Integral Test requires $f$ to be which on $[1,\infty)$?
Positive, continuous, decreasing — not necessarily a polynomial.
The Integral Test connects a series to an integral by setting $a_n=f(\ \_\_\ )$.
The terms are the function sampled at integers.
The conclusion
- $\displaystyle\sum a_n$ and $\displaystyle\int_1^\infty f(x)\,dx$ both converge or both diverge — same fate.
- Compute the improper integral: if it's finite, the series converges; if infinite, it diverges.
- (The series' sum is not the integral's value — they only share convergence, not the number.)
- One improper integral settles the whole series.
The Integral Test says $\sum a_n$ and $\int_1^\infty f\,dx$...
They share fate, not value.
The value of the improper integral equals the sum of the series.
They share convergence, not the value ($\sum\tfrac1{n^2}=\tfrac{\pi^2}{6}\neq1$).
Why it works
- The rectangles of area $a_n$ fit just above or below the curve $y=f(x)$.
- So the series is squeezed between $\int f$ and $\int f + a_1$ — finite integral means finite series.
- This is a rigorous "area comparison," turning a sum into an integral you can evaluate.
- It's especially handy for terms like $\tfrac1{n^p}$ and $\tfrac1{n\ln n}$.
Rectangles bounded by the curve
y = 1/x²
The term-rectangles fit against the area under $f$, so the series and the integral converge or diverge together.
Since $\int_1^\infty \tfrac1{x^2}\,dx=1$ (finite), the series $\sum\tfrac1{n^2}$...
Finite integral → convergent series.
If $\int_1^\infty f\,dx$ diverges (to $\infty$), then $\sum a_n$...
Same fate: divergent integral → divergent series.
The Integral Test needs $f$ positive, continuous, and decreasing — verify all three. And the test shares only convergence/divergence, not the value: $\sum a_n \neq \int_1^\infty f\,dx$ in general. Don't report the integral's value as the series' sum.
Test $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}$ with the Integral Test.
- $f(x)=\tfrac1{x^2}$ is positive, continuous, decreasing for $x\ge1$. ✓
- $\displaystyle\int_1^\infty \tfrac1{x^2}\,dx=1$ (finite, from lesson 6.13).
- The integral converges, so the series converges. (Its sum is $\tfrac{\pi^2}{6}$, not $1$.)
The Integral Test: if $f$ is positive, continuous, and decreasing with $a_n=f(n)$, then $\sum a_n$ and $\int_1^\infty f(x)\,dx$ share the same fate (both converge or both diverge). Evaluate the improper integral to decide — but its value is not the series' sum.