How to Use Comparison test for Convergence

• If \(0≤an≤bn\)​ for all \(n\) large enough and \(\sum b_n\) converges, then \(\sum a_n\) also converges.
• If \(an​≥bn​\) and \(\sum b_n\) diverges, then \(\sum a_n\) also diverges.

The Comparison Test decides a series by bounding it against one whose behavior is known: if a series is term-by-term smaller than a convergent series it converges, and if it is larger than a divergent series it diverges. Below: the conditions, and converging and diverging worked examples.

Here are two examples demonstrating the Comparison Test:

1. Converging Series Example:

Series: \(\sum_{n=1}^{\infty} \frac{2}{n^3 + 1}\)

Comparison Series: \(\sum_{n=1}^{\infty} \frac{1}{n^3}\)

• The terms \(\frac{2}{n^3 + 1}\) are smaller than \(\frac{2}{n^3}\) for all \(n \geq 1\), and we know that \(\sum \frac{1}{n^3}\) converges (since it’s a p-series with \(p = 3\)).
• By the Comparison Test, since the comparison series converges, the original series also converges.

2. Diverging Series Example:

Series: \(\sum_{n=1}^{\infty} \frac{1}{n \ln(n)}\)

Comparison Series: \(\sum_{n=1}^{\infty} \frac{1}{n}\)

• The series \(\frac{1}{n \ln(n)}\) is larger than \(\frac{1}{n^2}\) for large \(n\), and we know that the harmonic series \(\sum \frac{1}{n}\) diverges.
• By the Comparison Test, since the comparison series diverges, the original series diverges as well.

Frequently Asked Questions

What is the comparison test for convergence?

The comparison test compares a series with nonnegative terms to one whose behavior you already know. If 0 ≤ aₙ ≤ bₙ and ∑bₙ converges, then ∑aₙ converges. If aₙ ≥ bₙ ≥ 0 and ∑bₙ diverges, then ∑aₙ diverges. In short, smaller than a convergent series means convergent, and larger than a divergent series means divergent.

How do you use the comparison test for series?

Pick a simpler series with a known result, usually a p-series or geometric series, then show an inequality between the terms. For ∑ 1/(n² + 3), note that 1/(n² + 3) < 1/n² for every n ≥ 1. Since ∑ 1/n² is a p-series with p = 2 > 1, it converges, so the original series converges too.

When do you use the comparison test?

Use the comparison test when a series with positive terms looks like a simpler known series, for example when the terms are fractions with polynomials or roots, like 1/(n³ + 5) or 1/(√n − 0.5). It tells you whether the series converges or diverges, not its sum. If the inequality points the wrong way, try the limit comparison test instead.

Can you use the comparison test for alternating series?

Not directly, because the comparison test requires nonnegative terms. You can, however, apply it to the absolute values. For ∑ (−1)ⁿ/n², compare |aₙ| = 1/n² with itself or a larger convergent series; since ∑ 1/n² converges, the alternating series converges absolutely. To check a series like ∑ (−1)ⁿ/n, use the alternating series test.

What is the limit comparison test for convergence?

The limit comparison test says that if aₙ > 0, bₙ > 0 and lim aₙ/bₙ = c with 0 < c < ∞, then both series converge or both diverge. It avoids finding an exact inequality. For ∑ (3n + 1)/(n³ + 2), compare with 1/n²: the ratio approaches 3, and ∑ 1/n² converges, so the original series converges.

How do you use the comparison test for improper integrals?

For improper integrals, the idea is the same. If 0 ≤ f(x) ≤ g(x) for x ≥ a and ∫ g(x) dx converges, then ∫ f(x) dx converges. If f(x) ≥ g(x) ≥ 0 and ∫ g(x) dx diverges, so does ∫ f(x) dx. For example, 1/(x² + x) ≤ 1/x² for x ≥ 1, and ∫₁^∞ 1/x² dx = 1, so ∫₁^∞ 1/(x² + x) dx converges.

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