Calculus – Sequences and series – Comparison tests

Comparison tests (sometimes called: direct comparison tests)

Theorem: Suppose that \(\displaystyle\sum a_n\) and \(\displaystyle\sum b_n\) are series with positive terms.

1. If \(\displaystyle\sum b_n\) is convergent and \(a_n\leq b_n\) for all \(n\), then \(\displaystyle\sum a_n\) is also convergent.


2. If \(\displaystyle\sum b_n\) is divergent and \(a_n\geq b_n\) for all \(n\), then \(\displaystyle\sum a_n\) is also divergent.

Note that the conditions only need to hold for \(n\geq N\), where \(N\) is some fixed integer.

Examples:

1. \(\displaystyle\sum_{n=0}^{\infty}\frac{1}{2^n+1}\) is convergent, since \(\displaystyle\frac{1}{2^n+1}<\frac{1}{2^n}\) and \(\displaystyle\sum_{n=0}^{\infty}\frac{1}{2^n}\) is a geometric series with common ratio \(\displaystyle\frac{1}{2}\), which is convergent.


2. \(\displaystyle\sum_{n=0}^{\infty}\frac{1}{n^2+1}\) is convergent, since \(\displaystyle\frac{1}{n^2+1}<\frac{1}{n^2}\) and \(\displaystyle\sum_{n=0}^{\infty}\frac{1}{n^2}\) is a \(p\)-series with \(p=2>1\), which is convergent.


3. \(\displaystyle\sum_{n=1}^{\infty}\frac{\ln(n)}{n}\) is divergent, since \(\displaystyle\frac{\ln(n)}{n}>\frac{1}{n}\) for \(n\geq3\) and \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}\) is the harmonic series, which is divergent.

The limit comparison test

Theorem: Suppose that \(\displaystyle\sum a_n\) and \(\displaystyle\sum b_n\) are series with positive terms. If \(\displaystyle\lim\limits_{n\to\infty}\frac{a_n}{b_n}=c\), where \(c\) is a finite number and \(c>0\), then either both series converge or both diverge.

Proof:
Let \(m\) and \(M\) be positive numbers such that \(m < c < M\). Because \(\displaystyle\frac{a_n}{b_n}\) is close to \(c\) for large \(n\), there is an integer \(N\) such that

\[m < \frac{a_n}{b_n} < M\;\;\text{for}\;\;n>N\quad\Longrightarrow\quad mb_n < a_n < Mb_n\;\;\text{for}\;\;n>N.\]

If \(\displaystyle\sum b_n\) converges, so does \(\displaystyle\sum Mb_n\). Thus \(\displaystyle\sum a_n\) converges since \(a_n < Mb_n\) for \(n > N\).

If \(\displaystyle\sum b_n\) diverges, so does \(\displaystyle\sum mb_n\). Thus \(\displaystyle\sum a_n\) diverges since \(mb_n < a_n\) for \(n > N\).

Examples:

1) Consider the series \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^n-1}\). Let \(a_n=\displaystyle\frac{1}{2^n-1}\) and \(b_n=\displaystyle\frac{1}{2^n}\), then:

\[\lim\limits_{n\to\infty}\frac{a_n}{b_n}=\lim\limits_{n\to\infty}\frac{1}{2^n-1}\cdot\frac{2^n}{1}=\lim\limits_{n\to\infty}\frac{1}{1-2^{-n}}=\frac{1}{1-0}=1>0.\]

Since \(\displaystyle\sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}\frac{1}{2^n}\) is a geometric series with common ratio \(\displaystyle\frac{1}{2}\), which is convergent, we conclude that \(\displaystyle\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\frac{1}{2^n-1}\) is convergent too.

2) Consider the series \(\displaystyle\sum_{n=0}^{\infty}\frac{\sqrt{n^4+4n+5}}{2n^5+3n^2+1}\). Let \(a_n=\displaystyle\frac{\sqrt{n^4+4n+5}}{2n^5+3n^2+1}\) and \(b_n=\displaystyle\frac{\sqrt{n^4}}{n^5}=\frac{n^2}{n^5}=\frac{1}{n^3}\), then:

\[\lim\limits_{n\to\infty}\frac{a_n}{b_n}=\lim\limits_{n\to\infty}\frac{\sqrt{n^4+4n+5}}{2n^5+3n^2+1}\cdot\frac{n^3}{1}=\lim\limits_{n\to\infty}\frac{\sqrt{1+4n^{-3}+5n^{-4}}}{2+3n^{-3}+n^{-5}}=\frac{\sqrt{1+0+0}}{2+0+0}=\frac{1}{2}>0.\]

Since \(\displaystyle\sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}\frac{1}{n^3}\) is a \(p\)-series with \(p=3>1\), which is convergent, we conclude that \(\displaystyle\sum_{n=0}^{\infty}a_n=\sum_{n=0}^{\infty}\frac{\sqrt{n^4+4n+5}}{2n^5+3n^2+1}\) is convergent too.

3) Consider the series \(\displaystyle\sum_{n=0}^{\infty}\frac{3n^2+5n+2}{\sqrt{2n^5+4n^3+3}}\). Let \(a_n=\displaystyle\frac{3n^2+5n+2}{\sqrt{2n^5+4n^3+3}}\) and \(b_n=\displaystyle\frac{n^2}{\sqrt{n^5}}=\frac{n^2}{n^{\frac{5}{2}}}=\frac{1}{n^{\frac{1}{2}}}\), then:

\[\lim\limits_{n\to\infty}\frac{a_n}{b_n}=\lim\limits_{n\to\infty}\frac{3n^2+5n+2}{\sqrt{2n^5+4n^3+3}}\cdot\frac{n^{\frac{1}{2}}}{1}=\lim\limits_{n\to\infty}\frac{3+5n^{-1}+2n^{-2}}{\sqrt{2+4n^{-2}+3n^{-5}}}=\frac{3+0+0}{\sqrt{2+0+0}}=\frac{3}{\sqrt{2}}>0.\]

Since \(\displaystyle\sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}\frac{1}{n^{\frac{1}{2}}}\) is a \(p\)-series with \(p=\displaystyle\frac{1}{2}\leq1\), which is divergent, we conclude that \(\displaystyle\sum_{n=0}^{\infty}a_n=\sum_{n=0}^{\infty}\frac{3n^2+5n+2}{\sqrt{2n^5+4n^3+3}}\) is divergent too.

The ratio comparison test (not in the book)

Theorem: Suppose that \(\displaystyle\sum a_n\) and \(\displaystyle\sum b_n\) are series with positive terms.

1. If \(\displaystyle\sum b_n\) is convergent and \(\displaystyle\frac{a_{n+1}}{a_n}\leq\frac{b_{n+1}}{b_n}\) for all \(n\), then \(\displaystyle\sum a_n\) is also convergent.


2. If \(\displaystyle\sum b_n\) is divergent and \(\displaystyle\frac{a_{n+1}}{a_n}\geq\frac{b_{n+1}}{b_n}\) for all \(n\), then \(\displaystyle\sum a_n\) is also divergent.

Note that the conditions only need to hold for \(n\geq N\), where \(N\) is some fixed integer.

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Last modified on March 15, 2021