Comparison tests allow us to determine convergence or divergence of a series by comparing it to a series whose behavior we already know (like geometric or $p$-series). These tests are powerful tools for series with positive terms.
🎯 In this section you will learn
- The Direct Comparison Test and how to choose a comparison series
- The Limit Comparison Test (often easier to apply)
- When to use each test
📌 Direct Comparison Test
$$ \text{Suppose } 0 \le a_n \le b_n \text{ for all } n. $$
$$ \text{If } \sum b_n \text{ converges, then } \sum a_n \text{ converges.} $$
$$ \text{If } \sum a_n \text{ diverges, then } \sum b_n \text{ diverges.} $$
💡 Intuition
- If a larger series converges, a smaller series also converges.
- If a smaller series diverges, a larger series also diverges.
Determine if $\sum_{n=1}^\infty \frac{1}{n^2 + 1}$ converges.
①
Find a comparison series
$\frac{1}{n^2 + 1} \le \frac{1}{n^2}$ for all $n$.
②
Known series
$\sum \frac{1}{n^2}$ is a convergent $p$-series ($p=2>1$).
③
Apply Direct Comparison
Since $0 \le \frac{1}{n^2+1} \le \frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges, $\sum \frac{1}{n^2+1}$ converges.
Determine if $\sum_{n=1}^\infty \frac{1}{\sqrt{n+1}}$ diverges.
①
Find a comparison series
$\frac{1}{\sqrt{n+1}} \ge \frac{1}{\sqrt{2n}} = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{n}}$ for large $n$.
②
Known series
$\sum \frac{1}{\sqrt{n}} = \sum 1/n^{0.5}$ diverges ($p=0.5<1$).
③
Apply Direct Comparison
Since $\frac{1}{\sqrt{n+1}} \ge \frac{1}{\sqrt{2n}}$ and $\sum \frac{1}{\sqrt{n}}$ diverges, $\sum \frac{1}{\sqrt{n+1}}$ diverges.
📌 Limit Comparison Test
$$ \text{If } a_n, b_n > 0 \text{ and } \lim_{n \to \infty} \frac{a_n}{b_n} = L \text{ where } 0 < L < \infty, $$
$$ \text{then } \sum a_n \text{ and } \sum b_n \text{ either both converge or both diverge.} $$
💡 When to use Limit Comparison
The Limit Comparison Test is often easier when the inequality direction is unclear or when $a_n$ behaves like a simpler $b_n$ for large $n$.
Determine if $\sum_{n=1}^\infty \frac{2n^2 + 3n + 1}{n^4 + 5}$ converges.
①
Choose $b_n$
For large $n$, $a_n \approx \frac{2n^2}{n^4} = \frac{2}{n^2}$. Let $b_n = \frac{1}{n^2}$.
②
Compute the limit
$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{(2n^2+3n+1)/n^4+5}{1/n^2} = \lim_{n \to \infty} \frac{2n^2+3n+1}{n^4+5} \cdot n^2 = \lim_{n \to \infty} \frac{2n^4+3n^3+n^2}{n^4+5} = 2$
③
Conclusion
Since $0 < 2 < \infty$ and $\sum 1/n^2$ converges, $\sum a_n$ converges.
Determine if $\sum_{n=1}^\infty \frac{n}{n^2 + 1}$ converges or diverges.
①
Choose $b_n$
For large $n$, $a_n \approx \frac{n}{n^2} = \frac{1}{n}$. Let $b_n = \frac{1}{n}$.
②
Compute the limit
$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n/(n^2+1)}{1/n} = \lim_{n \to \infty} \frac{n^2}{n^2+1} = 1$
③
Conclusion
Since $\sum 1/n$ diverges, $\sum \frac{n}{n^2+1}$ diverges.
⚠️ Important Notes
- Direct Comparison requires inequality in the correct direction: To prove convergence, you need $a_n \le \text{convergent series}$. To prove divergence, you need $a_n \ge \text{divergent series}$.
- Limit Comparison only works for positive terms: The limit must be a positive finite number.
- Choosing the right $b_n$: Usually $b_n$ is the dominant term of $a_n$ simplified (e.g., $1/n^p$).
🔍 Key Takeaways
- Direct Comparison: Compare term-by-term to a known series.
- Limit Comparison: Compare limits of ratios; more flexible than Direct Comparison.
- Both tests require positive terms.
- Choose $b_n$ as a simpler series (geometric or $p$-series) that behaves like $a_n$ for large $n$.
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