Comparison Test/Limit Comparison Test for MATH 141
Exam Relevance for MATH 141
Comparison tests for series appear regularly in MATH 141. Useful when ratio/root tests are inconclusive.
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Understanding the Comparison Tests
Sometimes you'll encounter a series that doesn't fit neatly into the p-series or geometric series formulas, but it looks similar to one of those. The Comparison Test and Limit Comparison Test let you leverage what you already know — if your series behaves like a known series, it should converge or diverge like that series too.
The Direct Comparison Test
Setup: You have a series $\sum a_n$ and want to compare it to a known series $\sum b_n$.
Requirements: Both series must have positive terms (at least for $n \geq N$).
For Convergence (Squeeze Under a Convergent Series)
$$\text{If } 0 \leq a_n \leq b_n \text{ and } \sum b_n \text{ converges, then } \sum a_n \text{ converges}$$
Intuition: If your series is smaller than something finite, it must also be finite.
For Divergence (Squeeze Above a Divergent Series)
$$\text{If } 0 \leq b_n \leq a_n \text{ and } \sum b_n \text{ diverges, then } \sum a_n \text{ diverges}$$
Intuition: If your series is larger than something infinite, it must also be infinite.
The Limit Comparison Test
When direct comparison is tricky (inequalities are hard to establish), use the Limit Comparison Test.
Setup: Compare $\sum a_n$ to $\sum b_n$ where both have positive terms.
$$L = \lim_{n \to \infty} \frac{a_n}{b_n}$$
| Value of $L$ | Conclusion |
|---|---|
| $L > 0$ and finite | Both series converge or both diverge |
| $L = 0$ | If $\sum b_n$ converges, then $\sum a_n$ converges |
| $L = \infty$ | If $\sum b_n$ diverges, then $\sum a_n$ diverges |
The most common case: When $L$ is a positive finite number, the series behave the same way.
Choosing a Comparison Series
The key skill is picking a good series to compare to. Look at the dominant terms:
| Your Series Looks Like | Compare To | Known Behavior |
|---|---|---|
| $\frac{1}{n^p}$ terms | p-series $\sum \frac{1}{n^p}$ | Converges if $p > 1$ |
| $\frac{1}{2^n}$ or $r^n$ terms | Geometric $\sum r^n$ | Converges if $\|r\| < 1$ |
| $\frac{1}{n}$ in denominator | Harmonic $\sum \frac{1}{n}$ | Diverges |
Strategy: Simplify by keeping only the dominant (fastest-growing) terms in numerator and denominator.
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ converges or diverges.
Step 1: Find a comparison series
For all $n \geq 1$:
$$n^2 + 1 > n^2$$
$$\implies \frac{1}{n^2 + 1} < \frac{1}{n^2}$$
Step 2: Apply the Direct Comparison Test
So $0 < \frac{1}{n^2 + 1} < \frac{1}{n^2}$ for all $n \geq 1$.
Since $\sum \frac{1}{n^2}$ is a p-series with $p = 2 > 1$, it converges.
Step 3: Conclude
By the Direct Comparison Test, $\sum \frac{1}{n^2 + 1}$ converges.
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{\ln n}{n}$ converges or diverges.
Step 1: Find a comparison series
For $n \geq 3$: $\ln n > 1$, so:
$$\frac{\ln n}{n} > \frac{1}{n}$$
Step 2: Apply the Direct Comparison Test
We have $\frac{\ln n}{n} > \frac{1}{n}$ and $\sum \frac{1}{n}$ is the harmonic series, which diverges.
Step 3: Conclude
By the Direct Comparison Test, $\sum \frac{\ln n}{n}$ diverges.
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{n}{n^3 + 1}$ converges or diverges.
Step 1: Identify dominant terms
For large $n$: $\frac{n}{n^3 + 1} \approx \frac{n}{n^3} = \frac{1}{n^2}$
Step 2: Compute the limit
Compare to $b_n = \frac{1}{n^2}$:
$$L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{n}{n^3+1}}{\frac{1}{n^2}}$$
$$= \lim_{n \to \infty} \frac{n^3}{n^3+1}$$
$$= \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n^3}} = 1$$
Step 3: Apply the Limit Comparison Test
Since $L = 1$ (positive and finite), both series behave the same.
$\sum \frac{1}{n^2}$ converges (p-series, $p = 2 > 1$).
Therefore, $\sum \frac{n}{n^3 + 1}$ converges.
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ converges or diverges.
Step 1: Try direct comparison
We want to compare to $\frac{1}{2^n}$, but:
$$2^n - 1 < 2^n$$
$$\implies \frac{1}{2^n - 1} > \frac{1}{2^n}$$
This goes the wrong way for convergence! We'd need our series to be smaller than the convergent comparison.
Step 2: Use Limit Comparison instead
$$L = \lim_{n \to \infty} \frac{\frac{1}{2^n-1}}{\frac{1}{2^n}}$$
$$= \lim_{n \to \infty} \frac{2^n}{2^n - 1}$$
$$= \lim_{n \to \infty} \frac{1}{1 - \frac{1}{2^n}} = 1$$
Step 3: Conclude
Since $L = 1$ and $\sum \frac{1}{2^n}$ converges (geometric, $r = \frac{1}{2}$), the series $\sum \frac{1}{2^n - 1}$ converges.
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{3n^2 + 2n}{n^4 - n + 5}$ converges or diverges.
Step 1: Identify dominant terms
$$\frac{3n^2 + 2n}{n^4 - n + 5} \approx \frac{3n^2}{n^4} = \frac{3}{n^2}$$
Compare to $b_n = \frac{1}{n^2}$.
Step 2: Compute the limit
$$L = \lim_{n \to \infty} \frac{\frac{3n^2 + 2n}{n^4 - n + 5}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{(3n^2 + 2n) \cdot n^2}{n^4 - n + 5} = \lim_{n \to \infty} \frac{3n^4 + 2n^3}{n^4 - n + 5}$$
Divide top and bottom by $n^4$:
$$= \lim_{n \to \infty} \frac{3 + \frac{2}{n}}{1 - \frac{1}{n^3} + \frac{5}{n^4}} = \frac{3}{1} = 3$$
Step 3: Conclude
Since $L = 3$ (positive and finite) and $\sum \frac{1}{n^2}$ converges, the original series converges.
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{3^n}$ converges or diverges.
Step 1: Try Limit Comparison
For large $n$, exponentials dominate polynomials, so $\frac{n^2}{3^n}$ behaves like $\frac{1}{3^n}$.
Compare to $b_n = \frac{1}{3^n}$:
$$L = \lim_{n \to \infty} \frac{\frac{n^2}{3^n}}{\frac{1}{3^n}} = \lim_{n \to \infty} n^2 = \infty$$
This gives $L = \infty$! Since $\sum \frac{1}{3^n}$ converges, this case doesn't apply directly.
Step 2: Use Direct Comparison instead
For $n \geq 4$: $n^2 < 2^n$, so:
$$\frac{n^2}{3^n} < \frac{2^n}{3^n} = \left(\frac{2}{3}\right)^n$$
Step 3: Conclude
Since $\sum \left(\frac{2}{3}\right)^n$ converges (geometric with $r = \frac{2}{3} < 1$), the original series converges.
Quick Reference: Which Test to Use?
| Situation | Recommended Approach |
|---|---|
| Clear inequality (easy to see $a_n < b_n$) | Direct Comparison |
| Complicated expression | Limit Comparison |
| Polynomial ÷ Polynomial | Limit Comparison with p-series |
| Contains $2^n, 3^n$, etc. | Compare to geometric |
| Direct comparison goes wrong way | Switch to Limit Comparison |
Common Mistakes and Misunderstandings
❌ Mistake: Wrong direction for convergence
Wrong: "I showed $\frac{1}{n^2+1} > \frac{1}{n^3}$ and $\sum \frac{1}{n^3}$ converges, so my series converges."
Why it's wrong: For convergence, you need your series to be smaller than a convergent series, not larger. Being larger than something convergent tells you nothing.
Correct: Show $a_n \leq b_n$ where $\sum b_n$ converges.
❌ Mistake: Wrong direction for divergence
Wrong: "I showed $\frac{1}{n+1} < \frac{1}{n}$ and the harmonic series diverges, so my series diverges."
Why it's wrong: For divergence, you need your series to be larger than a divergent series. Being smaller than something divergent tells you nothing.
Correct: Show $a_n \geq b_n$ where $\sum b_n$ diverges.
❌ Mistake: Forgetting positive terms requirement
Wrong: Applying Comparison Test to $\sum \frac{(-1)^n}{n^2}$.
Why it's wrong: Both comparison tests require positive terms. Series with negative terms need different approaches (Alternating Series Test or test absolute values).
Correct: For series with mixed signs, consider testing $\sum |a_n|$ for absolute convergence first.
❌ Mistake: Bad limit in Limit Comparison
Wrong: Getting $L = 0$ or $L = \infty$ and concluding directly.
Why it's wrong: When $L = 0$, you can only conclude convergence if the comparison series converges. When $L = \infty$, you can only conclude divergence if the comparison series diverges. You need to check the condition!
Correct: If $L = 0$ or $L = \infty$, consider using a different comparison series or switching to Direct Comparison.
The Direct Comparison Test
Compare to a known series. For convergence: your series must be SMALLER than something convergent. For divergence: your series must be LARGER than something divergent.
Variables:
- $a_n$:
- terms of the series you're testing
- $b_n$:
- terms of the comparison series (known behavior)
The Limit Comparison Test
When L is positive and finite, both series behave the same way. This is the most commonly used case.
Variables:
- $a_n$:
- terms of the series you're testing (positive)
- $b_n$:
- terms of the comparison series (positive)
- $L$:
- the limit of the ratio
Dominant Term Strategy
For rational functions, the series behaves like 1/n^p where p is the difference in degrees. Keep only the highest-power terms to find your comparison series.
Variables:
- $p$:
- degree of denominator minus degree of numerator
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