Direct & Limit Comparison Tests

Core Concepts

Learn to determine a series' convergence by comparing it to a known series (like a p-series or geometric series). This stack covers both the Direct Comparison Test and the more flexible Limit Comparison Test.

Direct & Limit Comparison Tests

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What is the main strategy behind the Comparison Tests?

The strategy is to determine the convergence or divergence of an unknown series by comparing its terms to the terms of a known series (e.g., a p-series or a geometric series).

State the two conditions for the Direct Comparison Test (DCT) involving series \(\sum a_n\) and \(\sum b_n\).

1. If \(0 \le a_n \le b_n\) for all \(n\), and the 'larger' series \(\sum b_n\) converges, then the 'smaller' series \(\sum a_n\) also converges. 2. If \(0 \le a_n \le b_n\) for all \(n\), and the 'smaller' series \(\sum a_n\) diverges, then the 'larger' series \(\sum b_n\) also diverges.

When using the Direct Comparison Test, what happens if the 'larger' series diverges or the 'smaller' series converges?

The test is inconclusive in both cases. You learn nothing about the other series.

Use the Direct Comparison Test to determine if \(\sum_{n=1}^{\infty} \frac{1}{n^3+5}\) converges.

Compare it to \(\sum \frac{1}{n^3}\). We know \(n^3+5 > n^3\), so \(0 < \frac{1}{n^3+5} < \frac{1}{n^3}\). Since the 'larger' series \(\sum \frac{1}{n^3}\) is a convergent p-series (\(p=3 > 1\)), the 'smaller' series \(\sum \frac{1}{n^3+5}\) must also converge.

Use the Direct Comparison Test to determine if \(\sum_{n=2}^{\infty} \frac{1}{\sqrt{n}-1}\) diverges.

Compare it to \(\sum \frac{1}{\sqrt{n}}\). We know \(\sqrt{n}-1 < \sqrt{n}\), so \(\frac{1}{\sqrt{n}-1} > \frac{1}{\sqrt{n}}\). Since the 'smaller' series \(\sum \frac{1}{\sqrt{n}}\) is a divergent p-series (\(p=1/2 \le 1\)), the 'larger' series \(\sum \frac{1}{\sqrt{n}-1}\) must also diverge.

State the Limit Comparison Test (LCT) for two series \(\sum a_n\) and \(\sum b_n\) with positive terms.

Compute the limit \(L = \lim_{n \to \infty} \frac{a_n}{b_n}\). If \(L\) is a finite, positive number (\(0 < L < \infty\)), then both series either converge or both diverge.

In the LCT, what is the conclusion if \(L = \lim_{n \to \infty} \frac{a_n}{b_n} = 0\)?

If \(L=0\) and the comparison series \(\sum b_n\) converges, then the original series \(\sum a_n\) also converges.

In the LCT, what is the conclusion if \(L = \lim_{n \to \infty} \frac{a_n}{b_n} = \infty\)?

If \(L=\infty\) and the comparison series \(\sum b_n\) diverges, then the original series \(\sum a_n\) also diverges.

How do you typically choose the comparison series \(\sum b_n\) for the Limit Comparison Test when \(a_n\) is a rational function of \(n\)?

You create \(b_n\) by taking the term with the highest power of \(n\) in the numerator of \(a_n\) and dividing it by the term with the highest power of \(n\) in the denominator of \(a_n\).

Use the LCT on \(\sum_{n=1}^{\infty} \frac{2n^2+1}{5n^3+n}\). What is a good choice for \(\sum b_n\)?

Choose \(b_n = \frac{n^2}{n^3} = \frac{1}{n}\). Now compute \(\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{(2n^2+1)/(5n^3+n)}{1/n} = \frac{2}{5}\). Since the limit is finite and positive, and \(\sum \frac{1}{n}\) diverges (harmonic series), the original series also diverges.

When might the Limit Comparison Test be preferred over the Direct Comparison Test?

The LCT is often easier when the inequality for the DCT is difficult to establish or goes in the 'wrong' direction. For example, with \(\sum \frac{1}{n^2-1}\), direct comparison with \(\sum \frac{1}{n^2}\) fails because \(\frac{1}{n^2-1} > \frac{1}{n^2}\), but LCT works easily.

Direct & Limit Comparison Tests

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