Comparison Test
Suppose \(\sum_{n=1}^\infty a_n\) and \(\sum_{n=1}^\infty b_n\), are series with positive terms.
(i) if \(\sum_{n=1}^\infty b_n\) converges and \(a_n\leq b_n\) for every integer \(n\geq1\), then \(\sum_{n=1}^\infty a_n\) converges.
(ii) If \(\sum_{n=1}^\infty b_n\) diverges and \(a_n\geq b_n\) for every integer \(n\geq1\), then \(\sum_{n=1}^\infty a_n\) diverges.
Example
Determine whether each series converges or diverges.
(a) \(\sum_{n=1}^\infty\frac1{2+5^n}\)
(b) \(\sum_{n=1}^\infty\frac{3n}{\sqrt{n^3}+1}\)
Solution
(a) To find \(b_n\), focus on the denominator of \(a_n\) to construct an inequality that is true for all \(n\geq1\).
$$\begin{aligned}2+5^n&>5^n\\\frac{1}{5^n}&>\frac{1}{2+5^n}\end{aligned}$$
Take \(b_{n}=\frac{1}{5^{n}}\). The series \(\sum_{n=1}^{\infty}b_{n}=\sum_{n=1}^{\infty}\frac{1}{5^{n}}\) which is a geometric series with \(r=\frac{1}{5}<1\), and thus, converges.
By the Comparison Test Part (i), the series \(\sum_{n=1}^\infty\frac1{2+5^n}\) converges as well.
(b) to find \(b_n\), focus on both the numerator and the denominator of \(a_n\) to contruct an inequality that is true for all \(n\geq1\).
$$\frac{3n}{\sqrt{n^3}+1}>\frac{n}{n^{3/2}}=\frac{1}{n^{1/2}}$$
Take \(b_n=\frac{1}{n^{1/2}}\). The series \(\sum_{n=1}^{\infty}b_{n}=\sum_{n=1}^{\infty}\frac{1}{n^{1/2}}\) which is a p-series with \(p=\frac{1}{2}<1\), and thus, diverges.
By the Comparison Test Part (ii), the series \(\sum_{n=1}^\infty\frac{3n}{\sqrt{n^3}+1}\) diverges.
Limit Comparison Test
If \(\sum_{n=1}^\infty a_n\) and \(\sum_{n=1}^\infty b_n\) are series with positive terms and \(\underset{n\rightarrow\infty}{\operatorname*{lim}}\frac{a_{n}}{b_{n}}=c>0\), where \(c\) is a real number, then either both series converge or both diverge.
Note that the Limit Comparison Test is, in many cases, more convenient to use than the Comparison Test, as it does not require to create an inequality
Example
Determine whether each series converges or diverges.
(a) \(\sum_{n=1}^\infty\frac1{4n^2+1}\)
(b) \(\sum_{n=1}^\infty\frac1{\sqrt[3]{n^2+1}}\)
Solution
(a) Let \(a_n=\frac{1}{4n^2+1}\) and \(b_n=\frac1{n^2}\). Then,
$$\begin{aligned}
&\operatorname*{lim}_{x\to\infty}{\frac{a_{n}}{b_{n}}}=\operatorname*{lim}_{n\to\infty}{\frac{\frac{1}{4n^{2}+1}}{\frac{1}{n^{2}}}}=\operatorname*{lim}_{n\to\infty}{\frac{n^{2}}{4n^{2}+1}} \\
&=\lim_{n\to\infty}\frac{1}{4+1/n^{2}}=\frac{1}{4}>0
\end{aligned}$$
In this case, \(c=\frac{1}{4}>0\). Since \(\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty\frac1{n^2}\) is a p-series with \(p=2>1\), it converges, and therefore, the series \(\sum_{n=1}^\infty\frac1{4n^2+1}\) converges by the Limit Comparison Test.
(b) Let \(a_{n}=\frac{1}{\sqrt[3]{n^{2}+1}}\) and \(b_n=\frac1{n^{2/3}}\). Then,
$$\begin{aligned}
&\operatorname*{lim}_{x\to\infty}{\frac{a_{n}}{b_{n}}}=\operatorname*{lim}_{n\to\infty}{\frac{\frac{1}{\sqrt[3]{n^{2}+1}}}{\frac{1}{n^{2/3}}}}=\operatorname*{lim}_{n\to\infty}{\frac{n^{2/3}}{\sqrt[3]{n^{2}+1}}} \\
&=\lim_{n\to\infty}\frac{n^{2/3}}{\sqrt[3]{n^2(1+1/n^2)}}=\lim_{n\to\infty}\frac{n^{2/3}}{\sqrt[3]{n^2}\sqrt[3]{1+1/n^2}} \\
&=\lim_{n\to\infty}\frac{1}{\sqrt[3]{1+1/n^{2}}}=1>0
\end{aligned}$$
In this case, \(c=1>0\). Since \(\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty\frac1{n^{2/3}}\) is a p-series with \(p=\frac{2}{3}<1\), it diverges, and therefore, the series \(\sum_{n=1}^\infty\frac1{4n^2+1}\) diverges by the Limit Comparison Test.