9.3 The Comparison Test
Direct Comparison Test
Let $0<a_n \leq b_n$ for all $n$.
1. If $\sum_{n=1}^{\infty} b_n$ converges, then $\sum_{n=1}^{\infty} a_n$ converges.
2. If $\sum_{n=1}^{\infty} a_n$ diverges, then $\sum_{n=1}^{\infty} b_n$ diverges.
Limit Comparison Test
If $a_n>0, b_n>0$, and $\lim _{n \rightarrow \infty} \frac{a_n}{b_n}=L$, where $L$ is finite and positive, then both series either converge or both diverge.
Note: When choosing a series for comparison, you can disregard all but the highest powers of $n$ in both the numerator and denominator.
Example 1
- Determine whether the series is convergent or divergent.
(a) $\sum_{n=2}^{\infty} \frac{n}{n^2-3}$
(b) $\sum_{n=1}^{\infty} \frac{\sin ^2 n}{\sqrt{n^3}+1}$
▶️Answer/Explanation
Solution
(a) $\frac{n}{n^2-3}>\frac{n}{n^2}=\frac{1}{n}$ for all $n \geq 2$.
$\sum_{n=2}^{\infty} \frac{1}{n}$ is divergent harmonic series.
Therefore $\sum_{n=2}^{\infty} \frac{n}{n^2-3}$ is divergent by the Direct Comparison Test.
(b) $\frac{\sin ^2 n}{\sqrt{n^3}+1}=\frac{\sin ^2 n}{n^{3 / 2}+1} \leq \frac{1}{n^{3 / 2}+1}<\frac{1}{n^{3 / 2}}$ for $n \geq 1$.
$\sum_{n=1}^{\infty} \frac{1}{n^{3 / 2}}$ is convergent because it is a $p$ – series with $p=3 / 2>1$.
Therefore $\sum_{n=1}^{\infty} \frac{\sin ^2 n}{\sqrt{n^3}+1}$ is convergent by the Direct Comparison Test.
Example 2
- Determine whether the series is convergent or divergent.
(a) $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2+4}}$
(b) $\sum_{n=3}^{\infty} \frac{2^n}{3^n+1}$
▶️Answer/Explanation
Solution
(a) Let $a_n=\frac{1}{\sqrt{n^2+4}}$ and $b_n=\frac{1}{n}$.
$
\lim _{n \rightarrow \infty} \frac{a_n}{b_n}=\lim _{n \rightarrow \infty} \frac{1 / \sqrt{n^2+4}}{1 / n}=\lim _{n \rightarrow \infty} \frac{n}{\sqrt{n^2+4}}=1
$
Since $\sum b_n=\sum 1 / n$ is a divergent (harmonic series), the given series diverges by the Limit Comparison Test.
(b) Let $a_n=\frac{2^n}{3^n+1}$ and $b_n=\frac{2^n}{3^n}$
$
\lim _{n \rightarrow \infty} \frac{a_n}{b_n}=\lim _{n \rightarrow \infty} \frac{2^n /\left(3^n+1\right)}{2^n / 3^n}=\lim _{n \rightarrow \infty} \frac{3^n}{3^n+1}=1
$
Since $\sum b_n=\sum \frac{2^n}{3^n}=\sum\left(\frac{2}{3}\right)^n$ is a convergent geometric series, the given series converges by the Limit Comparison Test.
Example 3
- Which of the following series converge?
I. $\sum_{n=1}^{\infty} \frac{1}{n^2+n+3}$
II. $\sum_{n=1}^{\infty} \frac{\cos ^2 n}{n^2+2}$
III. $\sum_{n=1}^{\infty} \frac{1+4^n}{3^n}$
(A) I only
(B) II only
(C) III only
(D) I and II only
▶️Answer/Explanation
Ans:D
Example 4
- Which of the following series diverge?
I. $\sum_{n=1}^{\infty} \frac{1}{n !}$
II. $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}+2}$
III. $\sum_{n=1}^{\infty} \sin \left(\frac{1}{\mathrm{n}}\right)$
(A) I only
(B) II only
(C) II and III only
(D) I, II, and III
▶️Answer/Explanation
Ans:C
Example 5
- Which of the following series converge?
I. $\sum_{n=1}^{\infty} \frac{n^{3 / 2}}{3 n^3+7}$
II. $\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n^4+1}}$
III. $\sum_{n=1}^{\infty} \frac{n !}{n^n}$
(A) I only
(B) I and II only
(C) I and III only
(D) I, II, and III
▶️Answer/Explanation
Ans:D
Example 6
- Which of the following series cannot be shown to converge using the limit comparison test with the series $\sum_{n=1}^{\infty} \frac{1}{2^n}$ ?
(A) $\sum_{n=1}^{\infty} \frac{1}{2^n-1}$
(B) $\sum_{n=1}^{\infty} \frac{n}{2^n}$
(C) $\sum_{n=1}^{\infty} \frac{2 n}{2^{n+1} \sqrt{n^2+1}}$
(D) $\sum_{n=1}^{\infty} \frac{2 n^2-3 n}{2^n\left(n^2+n-100\right)}$
▶️Answer/Explanation
Ans:B