The Ratio Test

Most students assume the ratio test will settle every series they throw at it. It will not. The ratio test is a specialized tool: decisive for series built from factorials or exponentials, but useless when the limit equals 1. That limit, call it L, is what you compute from |a_n+1/a_n| as n goes to infinity. If L < 1, the series converges absolutely. If L > 1, the series diverges. If L = 1, the test says nothing, and you must reach for a different convergence test.

The ratio test, also called d'Alembert's ratio test, is stated formally in Stewart's Calculus, section 11.6. For a series Σ a_n, define L = lim_n→∞ |a_n+1/a_n|. Three outcomes are possible. L < 1 means the series converges absolutely. L > 1 means the series diverges. L = 1 means the test is inconclusive; the series may converge or diverge, and you need another method.

The ratio test requires that the limit of the ratio exists or is infinite. When the limit is infinite (L = ∞), the series diverges. That is the only case where the ratio test gives a decisive answer for a factorial-heavy series that grows faster than any geometric series.

The ratio test is your first move when the series contains factorials, constant powers like kⁿ, or nⁿ. These structures make the ratio a_n+1/a_n simplify cleanly, often to a constant or a fraction that depends only on n. For example, a series like Σ n!/kⁿ produces a ratio that cancels the factorial almost entirely, leaving a simple limit.

Avoid the ratio test for rational functions of n, such as Σ 1/n² or Σ (n²+1)/(n³+2). For those, the limit L will be 1, and the test will be inconclusive. Use the p-series test or comparison test instead. The ratio test also does not handle alternating series directly unless you take absolute values first, then apply the ratio test to the absolute series.

Three examples show the range of the ratio test: one where it converges, one where it diverges, and one where the factorial forces the ratio to infinity.

Example 1: Σ n/2ⁿ Converges

This series appears in Stewart's section 11.6. Let a_n = n/2ⁿ. Compute |a_n+1/a_n| = ((n+1)/2ⁿ⁺¹) / (n/2ⁿ) = (n+1)/(2n). As n → ∞, this ratio approaches 1/2. Since L = 1/2 < 1, the series converges absolutely.

Example 2: Σ n²/3ⁿ Converges

Paul's Online Math Notes uses this example. Let a_n = n²/3ⁿ. Then |a_n+1/a_n| = ((n+1)²/3ⁿ⁺¹) / (n²/3ⁿ) = (n+1)²/(3n²). The limit as n → ∞ is 1/3. L = 1/3 < 1, so the series converges.

Example 3: Σ n!/10ⁿ Diverges

Paul's Online Math Notes gives this. Let a_n = n!/10ⁿ. Then |a_n+1/a_n| = ((n+1)!/10ⁿ⁺¹) / (n!/10ⁿ) = (n+1)/10. As n → ∞, this ratio goes to infinity. L = ∞ > 1, so the series diverges.

Factorial ratios are the single place where students make algebra errors that kill the limit. Here is the rule that covers every ratio test problem with factorials:

(n+1)! = (n+1) × n!. When you write a_n+1/a_n, the factorial in the numerator (n+1)! cancels with the factorial in the denominator n! to leave just (n+1). For example, if a_n = n!/kⁿ, then a_n+1 = (n+1)!/kⁿ⁺¹, and the ratio becomes ((n+1)!/kⁿ⁺¹) × (kⁿ/n!) = (n+1)/k. The k comes from the exponential cancellation.

The same simplification works for any constant base. If a_n = (2n)!/nⁿ, then (2n+2)! = (2n+2)(2n+1)(2n)!, which still simplifies a_n+1/a_n to a polynomial ratio. Write the expansion once, cancel term by term, and the limit appears.

The ratio test fails when L = 1. This is not a rare edge case; it happens for every p-series Σ 1/n^p. For a_n = 1/n^p, the ratio |a_n+1/a_n| = (n/(n+1))^p. The limit is 1^p = 1 for any p. The ratio test says nothing about whether Σ 1/n² converges or Σ 1/n diverges.

When the ratio test is inconclusive, move to the p-series test (for rational powers), the comparison test, or the limit comparison test. The integral test also works for p-series, but it is slower on an exam. The root test is an alternative, but it will also give L = 1 for a p-series, so it is equally inconclusive.

A student who stops at L = 1 and guesses convergence or divergence has misapplied the test. The ratio test does not permit a conclusion at L = 1. That is the single most common error in using the ratio test.

When the ratio test returns L = 1, you have three options, in order of speed. First, check if the series is a p-series or geometric series. If it is, apply the p-series test or geometric series test directly. Second, try the limit comparison test against a known convergent or divergent series, usually a p-series. Third, use the integral test if the terms are positive and decreasing.

Do not reach for the root test as a backup. The root test gives the same limit for most series where the ratio test gives L = 1, especially rational functions and p-series. The root test is useful only when the terms contain nth powers, such as (1 + 1/n)ⁿ.

The conclusion must state three things: the value of L, the comparison to 1, and the verdict. For example, from Stewart's example Σ n/2ⁿ: "Since L = 1/2 < 1, the series converges absolutely by the ratio test." Do not write "converges" without specifying the test used, because the reader may wonder which test applies.

When L = ∞, write "Since L = ∞ > 1, the series diverges by the ratio test." When L = 1, write "The ratio test is inconclusive; the limit is 1." Then state the next test you will use. Never write "the series converges because the ratio test gives 1", that is false and will lose marks on an exam.

If you are using the ratio test as part of finding a power series radius of convergence, the conclusion is different: you compute R = 1/L (if L is finite and nonzero), then test endpoints separately. The power series convergence radius interval is determined by the ratio test, but the endpoints require a separate test such as the alternating series test or the p-series test.