Study Guide: Comparison Tests for Infinite Series

Learning Objectives

By the end of this study guide, you should be able to:

• Identify appropriate "reference series" (geometric or $p$-series) to compare against a given series.
• Apply the Direct Comparison Test (DCT) to prove convergence or divergence by establishing strict inequalities.
• Apply the Limit Comparison Test (LCT) when algebraic bounding is difficult but asymptotic behavior is clear.
• Recognize the limitations of comparison tests, specifically when terms are not strictly positive.

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Key Terms & Glossary

• Direct Comparison Test (DCT): A convergence test that directly compares the terms of an unknown series to a known series using inequalities ($<$, $>$, $\leq$, $\geq$).
• Limit Comparison Test (LCT): A convergence test that compares the asymptotic growth rates of two series by taking the limit of the ratio of their terms as $n \to \infty$.
• **$p-Series**: A series of the form \sum \frac{1}{n^p}. It acts as a standard benchmark, converging if p > 1$ and diverging if $p \leq 1$.
• Geometric Series: A series of the form $\sum a r^n. It converges when the common ratio |r| < 1$.
• Absolute Convergence: The property of a series where the sum of the absolute values of its terms converges. (Comparison tests only apply to non-negative terms; for series with negative terms, we test the absolute values).

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The "Big Idea"

[!TIP] The core principle is guilt by association.

When evaluating a complex infinite series, we strip away the "noise" (lower-order terms, constants) to reveal its fundamental behavior. By comparing this simplified version (our reference series) to the messy original, we can determine its fate.

If the messy series is always smaller than a series we know converges, the messy one must also converge (it's trapped). Conversely, if it is always larger than a series we know blows up to infinity, the messy one must also diverge.

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Formula / Concept Box

[!WARNING] The Limit Comparison Test provides no information if $L=0$ and $\sum b_n$ diverges, or if $L=\infty$ and $\sum b_n$ converges. You must pick a different comparison series!

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Hierarchical Outline

• 1. Foundations of Comparison
  • Both tests require non-negative terms: a_n \geq 0$, $b_n \geq 0.
  • If terms are negative, apply tests to $|a_n|$ to check for absolute convergence.
• 2. The Direct Comparison Test (DCT)
  • Requires establishing a term-by-term inequality.
  • Convergent Bounding: Must show $a_n \leq b_n$ (target is smaller than the convergent benchmark).
  • Divergent Bounding: Must show $a_n \geq b_n$ (target is larger than the divergent benchmark).
• 3. The Limit Comparison Test (LCT)
  • Bypasses the need for strict inequalities.
  • Computes ratio limit: $L = \lim_{n \to \infty} (a_n / b_n)$.
  • A finite, non-zero $L$ means the series grow at proportional rates.
• 4. Selecting Reference Series ($b_n$)
  • Keep highest powers of $n$ in the numerator and denominator.
  • Simplify to a basic $p$-series or geometric series.

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Visual Anchors

Choosing Your Test Strategy

Geometric Interpretation of Direct Comparison

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Definition-Example Pairs

• $p$-Series

  • Definition: A series whose terms are $1/n^p. Convergence depends entirely on the exponent p$.
  • Real-World Example: The Harmonic Series ($p=1$) models the harmonic overtones of a vibrating string. Even though terms shrink, the total sum diverges infinitely.

• Direct Comparison Test (DCT)

  • Definition: Proving convergence by showing a series is strictly smaller than a known convergent series.
  • Mathematical Example: Comparing $\sum \frac{1}{n^2+5}$ to $\sum \frac{1}{n^2}. Because adding 5 makes the denominator larger, the fraction is smaller. It is bounded by a convergent p$-series, so it converges.

• Limit Comparison Test (LCT)

  • Definition: Proving convergence by showing the ratio of a target series to a known series approaches a non-zero constant.
  • Mathematical Example: Testing $\sum \frac{1}{n^2 - 1}$. A direct inequality $1/(n^2-1) \leq 1/n^2$$ fails (it's actually larger). Instead, limit comparison with $1/n^2 yields $L=1$, proving convergence.

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Worked Examples

Example 1: Direct Comparison Test (Convergence)

Problem: Determine if $\sum_{n=1}^{\infty} \frac{1}{n^3 + 1}$ converges or diverges.

Step 1: Identify reference series. Drop the "+1$$. We are left with $$\sum \frac{1}{n^3}$$. Since p = 3 > 1, this reference series is a convergent p$-series.

Step 2: Establish inequality. For all positive integers $n$, $n^3 + 1 > n^3$. Therefore, taking the reciprocal reverses the inequality: $\frac{1}{n^3 + 1} < \frac{1}{n^3}$.

Step 3: Conclude. Since every term of our target series is smaller than a corresponding term of a known convergent series, $\sum \frac{1}{n^3 + 1}$ converges by the Direct Comparison Test.

Example 2: Limit Comparison Test

Problem: Determine if $\sum_{n=1}^{\infty} \frac{3n + 5}{\sqrt{n^4 + 1}}$ converges or diverges.

Step 1: Identify reference series. Look at highest powers. Numerator behaves like $n. Denominator behaves like$\sqrt{n^4} = n^2$. Our reference series$b_n = \frac{n}{n^2} = \frac{1}{n}$$. This is the harmonic series ($p=1$), which diverges.

Step 2: Set up the limit ratio. $L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \left( \frac{3n + 5}{\sqrt{n^4 + 1}} \cdot \frac{n}{1} \right)$ $L = \lim_{n \to \infty} \frac{3n^2 + 5n}{\sqrt{n^4 + 1}}$

Step 3: Evaluate limit. Divide top and bottom by $n^2$ (which is $\sqrt{n^4}$ inside the root): $L = \lim_{n \to \infty} \frac{3 + \frac{5}{n}}{\sqrt{1 + \frac{1}{n^4}}} = \frac{3 + 0}{\sqrt{1 + 0}} = 3$

Step 4: Conclude. Since $0 < L < \infty$, both series share the same fate. Because the reference series diverges, the target series diverges by the Limit Comparison Test.

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Checkpoint Questions

▶1. Why can't you use the Direct Comparison Test to prove that $\sum \frac{1}{n^2 - 1}$ converges by comparing it to $\frac{1}{n^2}$?

Because $n^2 - 1 < n^2, which means the reciprocal \frac{1}{n^2 - 1} > \frac{1}{n^2}$. The DCT requires the target series to be smaller than the convergent reference series, not larger. (You must use the Limit Comparison Test here).

▶2. If you use LCT and find that $L = 0$, what does this tell you?

It depends on your reference series $b_n$. If $b_n converges, then the target series a_n$ also converges (because $a_n is growing much slower than b_n$). If $b_n$ diverges, the test is inconclusive.

▶3. What must be true about the terms of a series before applying either the Direct or Limit Comparison Test?

The terms of both the target series and the reference series must be positive. If they alternate or have negative terms, you must test their absolute values instead.

▶4. What is the standard reference series for a target series of $a_n = \frac{1}{2^n - 1}$?

The geometric series $b_n = \frac{1}{2^n} = \left(\frac{1}{2}\right)^n$, which converges because $r = 1/2 < 1$.