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10.6 Comparison Tests for Convergence

Keywords

Comparison Test 直接比较判别法 • Limit Comparison Test 极限比较判别法

Benchmark Series 基准级数 • Dominant Terms 主导项

The Benchmarking Strategy

To use comparison tests, you must first have a "benchmark" series \(\sum b_n\) whose behavior is already known. Most often, we use:

  • \(p\)-series: \(\sum \frac{1}{n^p}\)
  • Geometric series: \(\sum ar^n\)

1. Direct Comparison Test (DCT)

The DCT is based on a simple logical "sandwich" idea. If all terms of your series are positive (\(a_n, b_n > 0\)):

  1. The Ceiling Rule: If \(a_n \le b_n\) and the "larger" series \(\sum b_n\) converges, then the "smaller" series \(\sum a_n\) must also converge.

  2. The Floor Rule: If \(a_n \ge b_n\) and the "smaller" series \(\sum b_n\) diverges, then the "larger" series \(\sum a_n\) must also diverge.

核心要点：Direct Comparison (LIM-7.A.8)

To use DCT effectively, you must establish a clear inequality (\(a_n \le b_n\) or \(a_n \ge b_n\)) for all \(n\) beyond a certain point.

2. Limit Comparison Test (LCT)

The DCT can be frustrating if the inequality goes the "wrong way." For example, \(\sum \frac{1}{n^2 - 1}\) is slightly larger than the convergent \(\sum \frac{1}{n^2}\), so the Ceiling Rule doesn't apply. This is where the Limit Comparison Test shines.

核心要点：Limit Comparison (LIM-7.A.9)

Suppose \(a_n > 0\) and \(b_n > 0\). Calculate the limit of the ratio of the two series:

\[ L = \lim_{n \to \infty} \frac{a_n}{b_n} \]

If \(0 < L < \infty\) (where \(L\) is a finite, positive number), then \(\sum a_n\) and \(\sum b_n\) share the same fate—they either both converge or both diverge.

How to Choose \(b_n\)?

Focus on the dominant terms (highest powers).

  • Example: For \(\sum \frac{3n + 5}{n^3 - 2n + 1}\), the dominant part is \(\frac{n}{n^3} = \frac{1}{n^2}\).
  • Compare it to the \(p\)-series \(b_n = \frac{1}{n^2}\) (which converges). Since the ratio limit will be \(3\), the original series also converges.

Visualizing Comparison

In the graph below, notice how \(a_n = \frac{1}{n^2+1}\) (blue) is always smaller than the benchmark \(b_n = \frac{1}{n^2}\) (orange). Since the area under \(1/n^2\) is finite, the area under the blue curve must also be finite.

Extension: Growth Hierarchy Revisited

When using LCT, the growth hierarchy we learned in Unit 1 is your best friend. \(\ln n \ll n^p \ll e^n\). If you are comparing \(\sum \frac{\ln n}{n^2}\) to \(\sum \frac{1}{n^{1.5}}\), knowing that \(n^{0.5}\) grows faster than \(\ln n\) tells you that \(\frac{\ln n}{n^2}\) will eventually be smaller than \(\frac{n^{0.5}}{n^2} = \frac{1}{n^{1.5}}\). This logic is the heart of "Asymptotic Analysis" in computer science.

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