Check the integral-test conditions

Rectangles that fit under a curve

The series \(\displaystyle\sum_{n=1}^{\infty}\frac{12}{(n+2)^2}\) has the terms \(\frac43\), \(\frac34\) and \(\frac{12}{25}\). Replace \(n\) by \(x\) to find the curve \(f(x)=\frac{12}{(x+2)^2}\), which passes through the point \((n,a_n)\) for every whole number \(n\).

Each teal rectangle is one unit wide and stands on the interval from \(n-1\) to \(n\). Its height is \(a_n\), shown by the orange dot at its upper right corner, so its area is the term \(a_n\).

Move the slider to add terms. Every rectangle lies under the curve, because the curve falls and each rectangle is as high as the curve at its right end.

The area of the rectangles is the sum of the terms, so it can be compared with the area under the curve. This needs a curve that lies above the axis, has no breaks and falls as \(x\) grows.

What the curve must do

Without the numbers, the curve is any function \(f\) with \(f(n)=a_n\). Each button breaks one property.

If part of the curve lies below the axis, the rectangles there have negative height and turn red. The function must be positive, which means that its curve lies above the axis.

A jump breaks the curve into two pieces. The function must be continuous, which means that its curve has no jumps or holes, so that the area under it is one integral.

On a rising part of the curve, a rectangle is higher than the curve and turns red. The function must be decreasing, which means that its curve falls as \(x\) grows.

The first terms do not decide convergence, so the three properties need to hold only for \(x\ge N\), where \(N\) is a whole number.

Notation

1. Replace \(n\) by \(x\) to write the term as a function of \(x\).

   \[\begin{aligned}a_n&=\frac{12}{(n+2)^2}\\ \cy{f(x)}&=\frac{12}{(x+2)^2}\end{aligned}\]

2. Positive means that the curve lies above the axis.

   \[f(x)>0\quad(x\ge N)\]

3. Continuous means that the curve has no jumps or holes. A fraction is continuous wherever its bottom is not 0.

   \[f\text{ is continuous on }[N,\infty)\]

4. Decreasing means that the curve falls as \(x\) grows. A negative slope \(f'(x)\) shows this.

   \[f'(x)<0\quad(x\ge N)\]

5. Choose the whole number \(N\) so that all three hold for \(x\ge N\).

When it fails

Add a wave: \(f(x)=\frac{12}{(x+2)^2}+b\sin^2(\pi x)\). At every whole number \(\sin(\pi x)=0\), so the rectangles and the terms do not change.

The curve stays positive and continuous, but it rises between the whole numbers, so it is not decreasing. Its area grows by \(\frac b2\) for each unit of \(x\) without limit, but the series stays the same.

Example 1: All three properties hold

Check the conditions of the integral test for \(\displaystyle\sum_{n=1}^{\infty}\frac{2}{(2n+3)^2}\).

1. Replace \(n\) by \(x\). At \(x=1\) the value is \(\frac2{25}\), which is the first term.

   \[\begin{aligned}f(x)&=\frac{2}{(2x+3)^2}\\ f(1)&=\tfrac2{25}\end{aligned}\]

2. For \(x\ge1\) the bottom is the square of a positive number, so \(f(x)>0\). The bottom is 0 only at \(x=-\frac32\), so \(f\) is continuous for \(x\ge1\).

3. Find the slope. It is negative for \(x\ge1\), because the top is negative and the bottom is positive. This means that \(f\) is decreasing.

   \[f'(x)=\frac{-8}{(2x+3)^3}<0\]

4. All three properties hold from \(N=1\), so the integral test can be used.

Example 2: A curve that rises first

Check the conditions of the integral test for \(\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^2}\).

1. Replace \(n\) by \(x\). For \(x>1\), \(\ln x\) is positive and \(x^2\) is not 0, so \(f\) is positive and continuous.

   \[f(x)=\frac{\ln x}{x^2}\]

2. A tempting wrong move: Compare \(f(1)=0\) with \(f(2)\approx0.17\). The second value is larger, so \(f\) does not decrease and the test cannot be used.

   \[f(1)=0

   The red bracket marks the rise, which ends near \(x=1.65\), and after it the curve falls. The test needs \(f\) to decrease only from some \(N\) on.

3. Find the slope. Its bottom \(x^3\) is positive, so it is negative when \(\ln x>\frac12\), which means \(x>\sqrt e\approx1.65\). The green label marks where the curve falls.

   \[\begin{aligned}f'(x)&=\frac{1-2\ln x}{x^3}\\ f'(x)<0&\iff x>\sqrt e\end{aligned}\]

4. The first whole number above 1.65 is 2, so all three properties hold from \(N=2\).

Practice

Check the conditions of the integral test for \(\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^2+5}\). Find the first whole number \(N\) from which \(f\) is decreasing.

Move the slider: the first rectangles grow before they shrink. Find where \(f'(x)\) is negative.

Questions

For question 1 the figure shows the curve \(f(x)=\frac{6x}{x^2+9}\) with the rectangles for \(a_n=f(n)\). For question 2 it shows the terms \(\frac1{n^2}\) and the curve \(g(x)=\frac1{x^2}+\frac12\sin^2(\pi x)\).

1. What is the first whole number \(n\) from which \(f\) is decreasing?

   • 3
   • 1
   • 2
   • 4
   Answer

   3

2. Why is \(g\) not suitable for the integral test?

   • It is not decreasing.
   • It is not positive.
   • It is not continuous.
   • It does not pass through the points \((n,\frac1{n^2})\).
   Answer

   It is not decreasing.

3. Which series has a function that is positive, continuous and decreasing for \(x\ge1\)?

   • \(\displaystyle\sum_{n=1}^{\infty}\frac{7}{n^2+2}\)
   • \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\)
   • \(\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{n+1}\)
   Answer

   \(\displaystyle\sum_{n=1}^{\infty}\frac{7}{n^2+2}\)

Summary

You write the term as a function \(f(x)\) and test that it is positive, continuous and decreasing from some \(N\) on.

These three properties put the rectangles under the curve, so the area under the curve can be compared with the sum of the terms.

Check these properties before every integral test, because the test says nothing when one of them fails.

This is one of four skills in topic 10.4 of Which & Why: AP Calculus.