Math Lesson 12.4.6 - The Ratio Convergence Test

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Welcome to our Math lesson on The Ratio Convergence Test, this is the sixth lesson of our suite of math lessons covering the topic of Infinite Series Explained, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Ratio Convergence Test Explained

You may not be very convinced with the reasoning used in the last three or four series about their convergence or divergence. Therefore, we use another rule about the convergence or divergence of a given series called the Ratio Test of Convergence. This test consists of calculating the limit of the ratio between two consecutive terms when the number of terms points to infinity. Here we will discuss the concept of limit very superficially, as much as we need to explain how to make the Ratio Convergence Test, as we will explain more extensively the concept of limit in the corresponding chapter. All what we can say here about this concept is that the limit of a function is the result obtained when the variable points towards a given number. For example, the limit of 3x + 5 when x points towards 12 is

Therefore, we say "when x points towards 12, then 3x + 5 points towards 41."

If we want to find the limit of a rational expression when the variable x point to infinity, we consider only the highest-order terms of the corresponding polynomials. In this case, we simplify the monomials obtained and if any variable is left after this simplification, we substitute its value by the infinity. This is because the value of a polynomial increases at a higher rate in the monomial with the highest degree, so the rest of terms are irrelevant. For example,

and so on.

But now, let's get back to the point, i.e. to the Ratio Test of Convergence. Let Σa_n be a series with positive terms, where a may be a variable, a monomial or an entire algebraic expression and let's suppose that

The Ratio Test of Convergence rule says that:

1. If L < 1, then Σa_n is a convergent series;
2. If L > 1, then Σa_n is a divergent series; and
3. If L = 1, then the convergence test of Σa_n is inconclusive, i.e. we need to use other methods to study its convergence.

In fact, we have silently used this ration test in previous cases in this tutorial. For example, in the p-series with p = 2, we have found a₁ = 1, a₂ = 1/4, a₃ = 1/9 and so on. Therefore, when we divide an + 1 by an, the result is always a number smaller than 1.

Example 4

Use the Ratio Test of Convergence to check the convergence of the following series.

1. Euler's Series
   e = ∞∑_{n = 1}1/n!
2. Harmonic series
   H = ∞∑_{n - 1}1/n
3. S = ∞∑_{n = 1}2ⁿ

Solution 4

1. We have for the Euler's Series
   a_n = 1/n! and a_{n + 1} = 1/(n + 1)!
   Hence,
   limn → ∞⁡a_{n + 1}/a_n = limn → ∞⁡1/(n + 1)!/1/n!
   = limn → ∞⁡1/(n + 1) ∙ n!/1/n!
   = limn → ∞⁡1/(n + 1) ∙ n! ∙ n!/1
   = limn → ∞⁡1/n + 1
   = 1/∞ + 1
   = 0
   Hence, since the limit L = 0, the Euler's series is convergent, from the first condition of the Ratio Comparison Test.
2. We have for the Harmonic Series a_n = 1/n and a_{n + 1} = 1/n + 1 Hence,
   limn → ∞⁡a_{n + 1}/a_n = limn → ∞⁡1/(n + 1)/1/n
   = limn → ∞⁡1/n + 1 ∙ n/1
   = limn → ∞⁡n/n + 1
   = limn → ∞ n/n
   = 1
   Since L = 1, the convergence of this series cannot be defined through the ratio convergence test. (Earlier we have said that this is a divergent series, but the proof is made with other methods).
3. We have for the series S
   a_n = 2ⁿ and a_{n + 1} = 2^{(n + 1)} Thus,
   limn → ∞⁡a_{n + 1}/a_n = limn → ∞⁡2^{(n + 1)}/2ⁿ
   = limn → ∞⁡2^{n + 1}/2ⁿ
   = limn → ∞⁡2ⁿ ∙ 2¹/2ⁿ
   = 2 ∙ limn → ∞⁡2ⁿ/2ⁿ
   = 2
   Thus, since the limit L is greater than 1, this series is diverging.

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