Section 5.4 Comparison Tests
In the previous section, we used comparisons of series with positive terms to improper integrals to do two things:
show that a series converges by comparing to a convergent improper integral, and
get an upper limit on the error in a partial sum as an approximation of the value of a series by comparing to an improper integral whose value we know.
We can do similar things by comparing one series to another instead of to an improper integral:
showing that a series converges by comparing to another series already known to converge, and
getting an upper limit on the error in a partial sum as an approximation of the value of a series by comparing to the known value of another series (often geometric!).
The basic idea is to compare a new series
\(\sum a_n\) to another series
\(\sum b_n\) whose convergence or divergence is already known:
Theorem 5.4.1. The [Basic] Comparison Test.
If sequences \(\{a_n\}\) and \(\{b_n\}\) have all terms non-negative and \(a_n \leq b_n\) for all terms, then Convergence of \(\sum b_n\) ensures convergence of \(\sum a_n\text{,}\) with sum no larger.
That is,
\begin{equation*}
\text{If } 0 \leq a_n \leq b_n \text{ for all } n \text{, then } 0 \leq \sum a_n \leq \sum b_n.
\end{equation*}
In the other direction, divergence of \(\sum a_n\) implies divergence of \(\sum b_n\text{.}\)
Common choices for the “known” series \(\sum b_n\) include geometric series \(\sum ar^n\) and \(p\)-series \(\ds\sum \frac{1}{n^p}\text{.}\)
Accuracy of partial sums by comparisons.
The above idea can also be applied to the remainder \(R_N = (\sum a_n) - S_N = \sum_{n=N+1}^\infty a_n\) to get an upper limit on the error in the partial sum \(S_N = \sum_{n=1}^N a_n\) and thus an upper bound on the exact value of the series:
If \(0 \leq a_n \leq b_n\) for \(n>N\) then
\begin{equation*}
0 \leq \left(\sum a_n\right) - S_N = \sum_{n=N+1}^\infty a_n \leq \sum _{n=N+1}^\infty b_n.
\end{equation*}
For example, if \(0 \leq a_n \leq ar^n\text{,}\) \(0 < r < 1\text{,}\) then
\begin{equation*}
0 \leq \left( \sum a_n \right) - S_N \leq \frac{ar^{N+1}}{1-r}.
\end{equation*}
Ignoring the first few terms and comparing with ratios.
Whether a series converges or not does not depend on the first few terms, even if “few” means “billion”: convergence only depends on what happens in the long run.
Loosely speaking: Do the terms individual terms \(a_n\) approach zero “fast enough”?
One useful measure of “fast enough” is at least as fast as a constant multiple of some convergent series:
The Limit Comparison Test.
If sequences \(\{a_n\}\) and \(\{b_n\}\) have all terms positive, and for some positive number \(c\)
\begin{equation*}
\lim_{n \to \infty} \frac{a_n}{b_n} = c, > 0,
\end{equation*}
then either both the series \(\sum a_n\) and \(\sum b_n\) converge, or both diverge.
Also, if \(a_n \leq c b_n\) for all \(n\) and some constant \(c\text{,}\) and \(\sum b_n\) converges, then \(\sum a_n\) converges too.
The Limit Comparison Test is often more convenient than the (basic) Comparison Test.
In particular, when a series has slightly larger terms than a similar convergent series, or slightly smaller terms than a similar divergent series, the inequalities are the wrong way around for the Comparison Test to say anything, but the similarities can be enough for Limit Comparison to work.
On the other hand, the Limit Comparison Test does not give an accuracy result for partial sums like the one given above for the original Comparison Test.
Study Guide.
Theorem 11 The Comparison Test, a.k.a. The Direct Comparison Test
Theorem 12 The Limit Comparison Test
Examples 17 and 18
Checkpoints 16 and 17
and one or several exercises from each of the groups 194–200 and 207–211.
openstax.org/books/calculus-volume-2/pages/5-4-comparison-tests
openstax.org/books/calculus-volume-2/pages/5-4-comparison-tests
