Comparison Tests for Convergence
| English | Chinese | Pinyin |
|---|---|---|
| comparison tests | 比较判别法 | bǐ jiào pàn bié fǎ |
Judging a series by a known neighbor
- If a mystery series looks like a series you already understand, compare them.
- The comparison tests 比较判别法 measure an unknown positive series against a known one (usually a p-series or geometric series).
- Bigger than something that diverges → diverges. Smaller than something that converges → converges.
- Two flavors: a direct comparison and a limit comparison.
Direct Comparison Test
- For positive terms with $0\le a_n\le b_n$:
- If $\sum b_n$ converges, then the smaller $\sum a_n$ converges too.
- If $\sum a_n$ diverges, then the bigger $\sum b_n$ diverges too.
- Squeeze from the right side: bounded above by a convergent series, or below by a divergent one.
Compare to a known series
A positive series bounded by a known convergent (or divergent) series shares its fate — pick the dominant comparison.
If $0\le a_n\le b_n$ and $\sum b_n$ converges, then $\sum a_n$...
Smaller than a convergent series → converges.
The comparison tests require positive terms.
They compare magnitudes of positive series.
Limit Comparison Test
- Sometimes inequalities are awkward; instead compare growth rates.
- Take $\displaystyle L=\lim_{n\to\infty}\dfrac{a_n}{b_n}$. If $L$ is a finite positive number ($0
- Pick $b_n$ from the dominant terms of $a_n$ (e.g. compare $\tfrac{n}{n^3+1}$ with $\tfrac1{n^2}$).
- The limit being positive and finite means they behave alike.
In the Limit Comparison Test, if $\lim\tfrac{a_n}{b_n}=L$ with $0
A finite positive limit → same convergence behavior.
For the Limit Comparison Test to conclude, the limit $L$ must be finite and ____.
$0
Choosing the comparison series
- Look at the dominant behavior of $a_n$ for large $n$ — the highest powers on top and bottom.
- $\dfrac{3n^2+1}{n^4+5}$ behaves like $\dfrac{3n^2}{n^4}=\dfrac{3}{n^2}$ — compare with the p-series $\tfrac1{n^2}$.
- Then either bound it directly or take the limit ratio.
- The right comparison series turns a scary series into a known one.
To test $\sum\dfrac{3n^2+1}{n^4+5}$, a good comparison series is...
Dominant behavior $\tfrac{3n^2}{n^4}=\tfrac{3}{n^2}$ → compare with $\tfrac1{n^2}$.
Since $\tfrac{1}{n^2+1}<\tfrac1{n^2}$ and $\sum\tfrac1{n^2}$ converges, $\sum\tfrac{1}{n^2+1}$...
Bounded above by a convergent series → converges.
Comparison tests need positive terms. For Direct Comparison, the inequality must point the right way: to prove convergence, bound above by a convergent series; to prove divergence, bound below by a divergent one. For Limit Comparison, the limit $L$ must be finite and positive ($0
Does $\displaystyle\sum\dfrac{1}{n^2+1}$ converge?
- For large $n$, $\dfrac{1}{n^2+1}$ behaves like $\dfrac{1}{n^2}$ (a convergent p-series).
- Direct: $\dfrac{1}{n^2+1}<\dfrac{1}{n^2}$, and $\sum\tfrac1{n^2}$ converges → so does the smaller series.
- Therefore $\sum\dfrac{1}{n^2+1}$ converges.
The comparison tests judge a positive series against a known one. Direct: bound above by a convergent series (→ converges) or below by a divergent one (→ diverges). Limit: if $\lim\frac{a_n}{b_n}$ is finite and positive, both share the same fate. Choose $b_n$ from $a_n$'s dominant terms.