Skip to main content

Section 9.5 Comparison, Ratio, and Root Tests

Subsection 9.5.1 Comparison Test

Suppose \(0 \leq a_n \leq b_n\) for all \(n\) beyond a certain point.

  • if \(\displaystyle \sum_{n = 1}^{\infty}b_n\) converges then \(\displaystyle \sum_{n = 1}^{\infty}a_n\) converges

  • if \(\displaystyle \sum_{n = 1}^{\infty}a_n\) diverges then \(\displaystyle \sum_{n = 1}^{\infty}b_n\) diverges

Example 9.5.1.

Given the series \(\displaystyle \sum_{n = 1}^{\infty} \dfrac{1}{n^{3}+1}\text{,}\) let's guess that it converges. We now need something bigger that converges.

Note 9.5.1.

TIP: \(\dfrac{\text{top smaller}}{\text{bottom bigger}} \lt \dfrac{\text{top}}{\text{bottom}} \lt \dfrac{\text{top bigger}}{\text{bottom smaller}}\)

One option is using the series \(\displaystyle \sum_{n = 1}^{\infty} \dfrac{1}{n^3}\text{.}\) We know this is a p-series so:

\begin{equation*} \boxed{ \sum_{n = 1}^{\infty} \dfrac{1}{n^3} \text{ converges becuase it is a p-series where } p = 3 \text{ and is bigger than} \sum_{n = 1}^{\infty} \dfrac{1}{n^{3}+1} \text{, so } \sum_{n = 1}^{\infty} \dfrac{1}{n^{3}+1} \text{ converges by the comparison test.} } \end{equation*}

Subsection 9.5.2 Limit Comparison Test

We really only care about series as \(n\) becomes large, so rather than a direct comparison, we can do a limit comparison.

Suppose \(a_n gt 0\) and \(b_n \gt 0\) for all \(n\) if

\begin{equation*} \lim_{n\to\infty} \dfrac{a_n}{b_n} = c \quad \text{where } c \gt 0 \end{equation*}

then both \(\displaystyle \sum a_n\) and \(\displaystyle \sum b_n\) converge or both diverge.

Example 9.5.2.

Determine if \(\displaystyle \sum_{n = 1}^{\infty} \dfrac{n^2 - 5}{n^3 + n + 2}\) converges or diverges:

\begin{equation*} a_n = \dfrac{n^2 - 5}{n^3 + n + 2}, \quad b_n = \dfrac{n^2}{n^3} = \dfrac{1}{2} \end{equation*}
\begin{align*} \lim_{n\to\infty} \dfrac{a_n}{b_n} \amp = \lim_{n\to\infty} \left( \dfrac{\frac{n^2 - 5}{n^3 + n + 2}}{\frac{n^2}{n^3}} \right) \\ \amp = \lim_{n\to\infty} \left( \dfrac{n^2 - 5}{n^3 + n + 2} \right) \left( \dfrac{n}{1} \right) \\ \amp = \lim_{n\to\infty} \dfrac{n^3 - 5n}{n^3 + n + 2} \\ \amp = 1 \gt 0 \end{align*}
\begin{equation*} \boxed{ \text{Since } \sum \frac{1}{n} \text{ diverges (harmonic series), } \sum \dfrac{n^2 - 5}{n^3 + n + 2} \text{ also diverges by the Limit Comparison Test.} } \end{equation*}

Subsection 9.5.3 Ratio Test

For a series \(\sum a_n\) suppose

\begin{equation*} \lim_{n\to\infty} \dfrac{\mid a_{n+1} \mid}{\mid a_n \mid} = L \end{equation*}

  • if \(L \lt 1\) then \(\sum a_n\) converges

  • if \(L \gt 1\) then \(\sum a_n\) diverges

  • if \(L = 1\) then the test is inconclusive

Example 9.5.3.

Determine if \(\displaystyle \sum_{n = 1}^{\infty} \dfrac{1}{n!}\) converges or diverges using the ratio test:

\begin{equation*} a_n = \dfrac{1}{n!}, \quad a_{n+1} = \dfrac{1}{(n+1)!} \end{equation*}
\begin{align*} \lim_{n\to\infty} \dfrac{ \mid \frac{1}{(n+1)!} \mid }{ \mid \frac{1}{n!} \mid} \amp = \lim_{n\to\infty} \dfrac{n!}{(n+1)!} \\ \amp = \lim_{n\to\infty} \dfrac{n!}{(n+1)n!} \\ \amp = \lim_{n\to\infty} \dfrac{1}{n+1}\\ \amp = 0 \lt 1 \end{align*}
\begin{equation*} \boxed{ \sum \dfrac{1}{n!} \text{ converges by the ratio test} } \end{equation*}

Subsection 9.5.4 Root Test

For a series \(\displaystyle \sum a_n\) with positive terms, suppose

\begin{equation*} \lim_{n\to\infty} \sqrt[n]{a_n} = \lim_{n\to\infty} (a_n)^{\frac{1}{n}} = L \end{equation*}

  • if \(L \lt 1\text{,}\) then \(\displaystyle \sum a_n\) converges

  • if \(L \gt 1\text{,}\) then \(\displaystyle \sum a_n\) diverges

  • if \(L = 1\text{,}\) then the test is inconclusive

Example 9.5.4.

Determine if \(\displaystyle \sum_{n=1}^{\infty} \left( \dfrac{4n-5}{2n+1} \right)^n\) converges or diverges:

\begin{align*} \lim_{n\to\infty} \sqrt[n]{\left( \dfrac{4n-5}{2n+1} \right)^n} \amp = \lim_{n\to\infty} \dfrac{4n-5}{2n+1} \\ \amp = 2 \gt 1 \end{align*}
\begin{equation*} \boxed{ \sum_{n=1}^{\infty} \left( \dfrac{4n-5}{2n+1} \right)^n \text{ diverges by the root test} } \end{equation*}