Use the integral to decide convergence
Rectangles that stay below a ceiling
The series \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^3}\) has the terms 1, \(\frac18\), \(\frac1{27}\) and so on. The curve \(f(x)=\frac{1}{x^3}\) passes through every point \((n,\frac{1}{n^3})\).
Each teal rectangle is one unit wide and as high as the curve at its right end, so its area is one term. The rectangles stand for the terms from \(n=2\) on, and they lie under the curve because the curve falls.
The terms from \(n=2\) on therefore add up to less than the area under the curve from 1 to infinity, which is \(\frac12\). A partial sum \(S_N\) is the sum of the first \(N\) terms, so with the first term, 1, every partial sum is below \(1+\frac12=\frac32\).
The terms are positive, so the partial sums rise. A list of numbers that rises but stays below a fixed number approaches a limit, so the series converges.
Two sets of rectangles trap the partial sums
Without the numbers, the curve is any positive, continuous, decreasing function \(f\). Slide \(w\) from 1 to 0. The orange corner of each rectangle moves from its right end to its left end.
At \(w=1\), the rectangles lie under the curve, so their total is at most the area, and the partial sums have an upper bound.
At \(w=0\), the rectangles lie above the curve, so their total is at least the area, and the partial sums have a lower bound.
If the area is finite, the partial sums have an upper bound, so the series converges. If the area is infinite, they have no upper bound, so the series diverges.
Notation
The two sets of rectangles give a lower bound and an upper bound for the partial sum \(S_N\).
\[\begin{aligned}\cb{\int_1^{N+1}f\,dx}&\le S_N\\ S_N&\le a_1+\cb{\int_1^{N}f\,dx}\end{aligned}\]Let \(a_n=f(n)\), where \(f\) is positive, continuous and decreasing for \(x\ge1\). The integral test says that the series and the integral both converge or both diverge.
\[\begin{aligned}&\sum_{n=1}^{\infty}a_n\text{ converges}\\ &\quad\iff\int_1^{\infty}f(x)\,dx\text{ converges}\end{aligned}\]If the integral is a number, write that the series converges by the integral test. If the integral is infinite, write that the series diverges by the integral test.
The test shows only whether the sum is finite. It does not give the sum, and the value of the integral is in general a different number.
\[\begin{aligned}\sum_{n=1}^{\infty}a_n&\ne\int_1^{\infty}f(x)\,dx\\ &\text{(in general)}\end{aligned}\]
When it fails
For \(\sum\frac{1}{n^3}\), the integral from 1 to infinity is \(\frac12\), but the sum is about 1.202.
The first term alone is 1, which is already more than the integral. Both numbers are finite, but they are not equal.
Example 1: The integral is infinite
Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{2n+3}\) converges.
Take \(f(x)=\dfrac{1}{2x+3}\). It is positive and continuous for \(x\ge1\). Its slope is negative, so it is decreasing.
\[f'(x)=\frac{-2}{(2x+3)^2}<0\]Find the area from 1 to \(b\), and let \(b\) grow.
\[\int_1^b\frac{dx}{2x+3}=\tfrac12\ln\frac{2b+3}{5}\to\infty\]The integral is infinite, so the series diverges by the integral test. The rectangles lie above the curve, so the partial sums also grow without bound.
Example 2: The integral is finite
Decide whether \(\displaystyle\sum_{n=1}^{\infty}e^{-n/2}\) converges. Then state what the integral tells you about the sum.
Take \(f(x)=e^{-x/2}\). It is positive and continuous, and its slope \(-\frac12e^{-x/2}\) is negative, so it is decreasing. Find the area from 1 to \(b\).
\[\begin{aligned}\int_1^b e^{-x/2}\,dx&=2e^{-1/2}-2e^{-b/2}\\ &\to\cb{2e^{-1/2}}\end{aligned}\]A tempting wrong move: The integral is \(2e^{-1/2}\approx1.21\), so the sum of the series is 1.21.
\[\sum_{n=1}^{\infty}e^{-n/2}=2e^{-1/2}\]The red parts of the rectangles lie above the curve, so the sum is more than the area, 1.21. The sum is at most the first term plus the area, about \(0.61+1.21=1.82\). As a check, this geometric series has the sum 1.54.
The integral is a number, so the series converges by the integral test, but the integral does not give the sum.
Practice
Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2+2n+5}\) converges.
Write the denominator as \((x+1)^2+4\). An antiderivative of \(\dfrac{1}{u^2+4}\) is an arctangent.
Solution
Take \(f(x)=\dfrac{1}{(x+1)^2+4}\). It is positive and continuous. Its slope is negative for \(x\ge1\), so it is decreasing.
\[f'(x)=\frac{-2(x+1)}{\bigl((x+1)^2+4\bigr)^2}<0\]Find the area from 1 to \(b\). An antiderivative is \(\tfrac12\arctan\dfrac{x+1}{2}\).
\[\begin{aligned}&\int_1^b f(x)\,dx\\ &=\tfrac12\left(\arctan\tfrac{b+1}{2}-\tfrac{\pi}{4}\right)\end{aligned}\]As \(b\) grows, the arctangent goes to \(\frac{\pi}{2}\), so the integral is \(\frac{\pi}{8}\), a number, and the series converges. Its sum is more than \(\frac{\pi}{8}\), because the rectangles lie above the curve.
\[\lim_{b\to\infty}\int_1^b f(x)\,dx=\cb{\tfrac{\pi}{8}}\]
Questions
The figure shows \(f(x)=\dfrac{1}{x+1}\) and the rectangles for the series \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n+1}\). The area under the curve from 1 to \(b\) is \(\ln\dfrac{b+1}{2}\), which grows past every number as \(b\) grows.
How does the total area of the rectangles compare with the area under the curve over the same interval?
- It is larger, because the rectangles stick out above the curve.
- It is smaller, because the rectangles lie under the curve.
- It is equal, because the rectangles touch the curve.
Answer
It is larger, because the rectangles stick out above the curve.
The rectangles lie above the curve, and the area under the curve has no upper bound. What follows for the series?
- It diverges, because its partial sums are at least as large as the area.
- It converges, because the terms \(\frac{1}{n+1}\) approach 0.
- It converges to the value of the integral.
Answer
It diverges, because its partial sums are at least as large as the area.
What does the integral test show for \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{(n+1)^{3/2}}\)?
- It converges, because the integral is \(\sqrt2\), a number.
- It converges to \(\sqrt2\).
- It diverges, because the integral is positive.
Answer
It converges, because the integral is \(\sqrt2\), a number.
Summary
You find the improper integral, and the series converges if the integral is a number and diverges if it is infinite.
This works because rectangles under the curve bound the partial sums from above, and rectangles above the curve bound them from below.
It applies when the terms come from a positive, decreasing function you can integrate, and it shows only whether the sum is finite.
This is one of four skills in topic 10.4 of Which & Why: AP Calculus.