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Calculus BC · Unit 10 · 10.4
AP Calculus BC · Infinite Sequences and Series
The Integral Test for Convergence
Rectangles against area: the sum and the improper integral share a fate, never a value.
key idea$\sum_{n=1}^{\infty}a_n\ \text{and}\ \int_{1}^{\infty}f(x)\,dx\ \text{share a fate},\ \ a_n=f(n)$
The test
Applies to
$a_n=f(n)$ where $f$ is positive, continuous and decreasing on $[k,\infty)$
Conclusion
the series and the improper integral both converge or both diverge
Hypotheses
all three must be checked and stated; they are part of the answer
Best for
terms containing $\ln n$, or any $f$ you can actually antidifferentiate
Never gives
the sum of the series — the integral's value is not the series' value
Compare a sum of rectangles with the area under a curve. If the curve encloses finite area the rectangles must too, and if the area is infinite so is the sum. The three hypotheses are what make that rectangle picture valid, so they are not optional bookkeeping.
| Series | $f(x)$ | $\displaystyle\int_{1}^{\infty}f$ | Verdict |
|---|
| $\displaystyle\sum\frac{1}{n^{2}}$ | $x^{-2}$ | $=1$ | converges |
| $\displaystyle\sum\frac{1}{n}$ | $x^{-1}$ | diverges | diverges |
| $\displaystyle\sum\frac{1}{n\ln n}$ | $\dfrac{1}{x\ln x}$ | diverges | diverges |
| $\displaystyle\sum\frac{1}{n(\ln n)^{2}}$ | $\dfrac{1}{x(\ln x)^{2}}$ | $=\dfrac{1}{\ln2}$ | converges |
| $\displaystyle\sum ne^{-n^{2}}$ | $xe^{-x^{2}}$ | $=\dfrac{1}{2e}$ | converges |
Rows 3 and 4 are the pair this test exists for. They differ by a single exponent and land on opposite sides — no comparison or ratio test separates them, but the integral does it in one line.
Positive$f(x)\gt 0$ on the interval
Continuousno breaks past the start
Decreasingcheck $f'(x)\lt 0$
When to reach for it
| Reach for the integral test when… | Reason |
|---|
| the term contains $\ln n$ | logarithms respond well to substitution in the integral |
| $f$ has an obvious antiderivative | the whole test is one improper integral |
| comparison candidates are ambiguous | the integral decides borderline cases exactly |
| the term involves $e^{-n^{2}}$ or similar | a $u$-substitution collapses it |
| not when terms alternate | the hypotheses require $f\gt 0$; use the alternating series test |
If the antiderivative is out of reach, so is this test. Check that you can actually integrate $f$ before committing — otherwise a comparison is the faster route.
Worked examples
EX 1
Verify the hypotheses first
$\displaystyle\sum\frac{1}{n^{2}+1}$Determine whether $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^{2}+1}$ converges.
1
Define $f$ and check positivity
$f(x)=\dfrac{1}{x^{2}+1}\gt 0$ for all $x\ge1$. ✓
2
Check continuity and monotonicity
The denominator never vanishes, so $f$ is continuous; $f'(x)=\dfrac{-2x}{\left(x^{2}+1\right)^{2}}\lt 0$, so $f$ is decreasing. ✓
3
Evaluate the improper integral
$\displaystyle\int_{1}^{\infty}\dfrac{dx}{x^{2}+1}=\lim_{b\to\infty}\left[\arctan x\right]_{1}^{b}=\dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}$.
4
Conclude
The integral converges, so the series converges.
5
Resist a false step
The sum is not $\dfrac{\pi}{4}$. The integral shares the series' fate, not its value.
EX 2
The logarithm case
substitution does the workDetermine whether $\displaystyle\sum_{n=2}^{\infty}\dfrac{1}{n\ln n}$ converges.
1
Check the hypotheses on $[2,\infty)$
$f(x)=\dfrac{1}{x\ln x}$ is positive, continuous and decreasing there. ✓
2
Substitute
$u=\ln x$, $du=\dfrac{dx}{x}$, turning the integral into $\displaystyle\int_{\ln2}^{\infty}\dfrac{du}{u}$.
3
Evaluate
$\displaystyle\lim_{b\to\infty}\left[\ln u\right]_{\ln2}^{b}=\infty$.
4
Conclude
The integral diverges, so the series diverges.
5
Note how narrowly
The terms shrink faster than $\dfrac1n$ yet still not fast enough. This is the boundary case that motivates the whole test.
EX 3
One exponent later
the other side of the boundaryDetermine whether $\displaystyle\sum_{n=2}^{\infty}\dfrac{1}{n(\ln n)^{2}}$ converges.
1
Use the same substitution
$u=\ln x$ gives $\displaystyle\int_{\ln2}^{\infty}\dfrac{du}{u^{2}}$.
2
Evaluate
$\left[-\dfrac1u\right]_{\ln2}^{\infty}=\dfrac{1}{\ln2}$, a finite number.
3
Conclude
The integral converges, so the series converges.
4
Compare with the previous example
Squaring the logarithm was enough to flip the verdict.
5
Draw the moral
Series near this boundary cannot be settled by intuition. The integral test is the instrument precise enough to separate them.
EX 4
Checking ‘decreasing’ properly
it can fail at firstFor $\displaystyle\sum_{n=1}^{\infty}\dfrac{\ln n}{n^{2}}$, show the hypotheses hold and determine convergence.
1
Differentiate to test monotonicity
$f(x)=\dfrac{\ln x}{x^{2}}$ gives $f'(x)=\dfrac{1-2\ln x}{x^{3}}$.
2
Find where it turns
$f'\lt 0$ when $\ln x\gt\dfrac12$, that is $x\gt\sqrt{e}\approx1.65$. So $f$ increases at first.
3
Start the test later
Apply the test on $[2,\infty)$, where all three hypotheses hold. Dropping finitely many terms cannot change convergence.
4
Integrate by parts
$\displaystyle\int_{2}^{\infty}\dfrac{\ln x}{x^{2}}dx=\left[-\dfrac{\ln x}{x}-\dfrac1x\right]_{2}^{\infty}=\dfrac{\ln2+1}{2}$, finite.
5
Conclude
The series converges. Shifting the starting point is the standard repair when monotonicity fails early.
What it cannot tell you
The integral's value is not the sum
They share a verdict, not a number. $\sum\dfrac{1}{n^{2}}=\dfrac{\pi^{2}}{6}$ while the matching integral equals $1$.
All three hypotheses must be stated
Positive, continuous, decreasing. On free response these are scored, and omitting them costs the point even with the right verdict.
Monotonicity may start late
If $f$ increases for small $x$, begin the test at a larger $k$. Finitely many terms never affect convergence.
It cannot handle alternating series
The hypotheses demand $f\gt 0$. Terms with $(-1)^{n}$ need the alternating series test instead.