9.2 The Integral Test and $p$-Series
The Integral Test
If $f$ is positive, continuous, and decreasing on $[1, \infty)$ and $a_n=f(n)$, then
$
\sum_{n=1}^{\infty} a_n \quad \text { and } \quad \int_1^{\infty} f(x) d x
$
either both converge or both diverge. In other words:
1. If $\sum_{n=1}^{\infty} a_n$ is convergent, then $\int_1^{\infty} f(x) d x$ is convergent.
2. If $\sum_{=1}^{\infty} a_n$ is divergent, then $\int_1^{\infty} f(x) d x$ is divergent.
Example1
- Determine whether the series is convergent or divergent.
(a) $\sum_{n=1}^{\infty} \frac{1}{n^2+1}$ (b) $\sum_{n=1}^{\infty} \frac{\ln n}{n}$
▶️Answer/Explanation
Solution
(a)
The function $f(x)=1 /\left(x^2+1\right)$ is continuous, positive, and decreasing on $[1, \infty)$ so we can use the Integral Test:
$
\begin{aligned}
& \int_1^{\infty} \frac{1}{x^2+1} d x=\lim _{b \rightarrow \infty} \int_1^b \frac{1}{x^2+1} d x=\lim _{b \rightarrow \infty}\left[\tan ^{-1} x\right]_1^b \\
& =\lim _{b \rightarrow \infty}\left[\tan ^{-1} b-\tan ^{-1} 1\right]=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}
\end{aligned}
$
So the series converges.
Note: The fact that the integral converges to $\pi / 4$ does not imply that the infinite series converges to $\pi / 4$.
(b)
The function $f(x)=\ln x / x$ is continuous, positive, and decreasing on $[1, \infty)$ so we can use the Integral Test:
$
\begin{aligned}
& \int_1^{\infty} \frac{\ln x}{x} d x=\lim _{b \rightarrow \infty} \int_1^b \frac{\ln x}{x} d x=\lim _{b \rightarrow \infty}\left[\frac{\ln ^2 x}{2}\right]_1^b \\
& =\lim _{b \rightarrow \infty}\left[\frac{\ln ^2 b}{2}-\frac{\ln ^2 1}{2}\right]=\infty-0=\infty
\end{aligned}
$
So the series diverges.
p- Series and Harmonic Series
The $p$ – series $\sum_{n=1}^{\infty} \frac{1}{n^p}=\frac{1}{1^p}+\frac{1}{2^p}+\frac{1}{3^p}+\frac{1}{4^p}+\cdots$ is convergent if $p>1$ and divergent if $0<p \leq 1$.
For $p=1$, the series $\sum_{n=1}^{\infty} \frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$ is called harmonic series. A general harmonic series is of the form $\sum \frac{1}{(\boldsymbol{a n}+\boldsymbol{b})}$.
Example 2
- Determine whether the series is convergent or divergent
(a) $1+\frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{9}}+\frac{1}{\sqrt[3]{16}}+\frac{1}{\sqrt[3]{25}}+\cdots$
(b) $\sum_{n=1}^{\infty} n^{1-\pi}$
▶️Answer/Explanation
Solution
(a) $1+\frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{9}}+\frac{1}{\sqrt[3]{16}}+\frac{1}{\sqrt[3]{25}}+\cdots=\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n^2}}=\sum_{n=1}^{\infty} \frac{1}{n^{2 / 3}}$
The $p$ – series is divergent since $p=2 / 3<1$.
(b) $\sum_{n=1}^{\infty} n^{1-\pi}=\sum_{n=1}^{\infty} n^{-(\pi-1)}=\sum_{n=1}^{\infty} \frac{1}{n^{\pi-1}}$
The $p$ – series is convergent since $p=\pi-1 \approx 2.14>1$.
Example 3
- If $\int_1^{\infty} \frac{d x}{x^2+1}=\frac{\pi}{4}$, then which of the following must be true?
I. $\sum_{n=1}^{\infty} \frac{1}{n^2+1}$ diverges.
II. $\sum_{n=1}^{\infty} \frac{1}{n^2+1}$ converges.
III. $\sum_{n=1}^{\infty} \frac{1}{n^2+1}=\frac{\pi}{4}$
(A) none
(B) I only
(C) II only
(D) II and III only
▶️Answer/Explanation
Ans:C
Example 4
- What are all values of $p$ for which $\int_1^{\infty} \frac{1}{\sqrt[3]{x^p}}$ converges?
(A) $P<-3$
(B) $P<-1$
(C) $P>1$
(D) $P>3$
▶️Answer/Explanation
Ans:D
Example 5
- Which of the following series converge?
I. $\sum_{n=1}^{\infty} \frac{n}{2 n^2+1}$
II. $\sum_{n=1}^{\infty} n e^{-n^2}$
III. $\sum_{n=2}^{\infty} \frac{1}{x \ln x}$
(A) I only
(B) II only
(C) III only
(D) I and II only
▶️Answer/Explanation
Ans:B
Example 6
- Which of the following series converge?
I. $\sum_{n=1}^{\infty} \frac{n}{2 n^2+1}$
II. $\sum_{n=1}^{\infty} n e^{-n^2}$
III. $\sum_{n=2}^{\infty} \frac{1}{x \ln x}$
(A) I only
(B) II only
(C) III only
(D) I and II only
▶️Answer/Explanation
Ans:D