Math Lesson 12.4.4 - The Comparison Test of Convergence

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Welcome to our Math lesson on The Comparison Test of Convergence, this is the fourth 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.

Understanding the Comparison Test of Convergence

Let's consider two infinite series with non-negative term

∞∑_{n = 1}x_n and ∞∑_{n = 1}y_n

If x_n ≤ y_n for sufficiently large values of n, then

1. If ∞∑_{n = 1}y_n converges, then ∞∑_{n = 1}x_n converges too, and
2. If ∞∑_{n = 1}x_n diverges, then ∞∑_{n = 1}y_n diverges too.

The above rule is not only true for geometric series but for all infinite series. However, for now we will illustrate this convergence test with infinite geometric series only, as we have not explained yet the other infinite series.

For example, if we consider the infinite geometric series

it is easy to see that this is a converging series, as the first terms are

and so on. Therefore, it is easy to see that this series is identical to that discussed in the first example in theory (the one with the figure), the sum of which, was equal to 1.

Therefore, the other infinite geometric series

is also convergent, as

for any positive integer n.

Indeed,

Thus, we have x₁ = 1/3 and R = 1/3 too. Hence, from the formula for the calculation of the sum of an infinite geometric series, we have

Let's consider now two diverging geometric series

where the general term of S₁(n) is x_n = 4 · 2n and that of S₂(n) is yn = 5 · 3n.

The first series is diverging, as the more the value of n increases, the more S₁(n) increases too. Therefore, this is a diverging geometric series, as when n points to infinity, so does S₁(n) as well.

Obviously, when comparing the general terms we obtain

for all positive integers n. Hence, if we prove that S₁(n) is divergent, it is not necessary to prove that S₂(n) is divergent, as this is implied from the above rule. Let's check the convergence of S₁(n). In simple words, a series is convergent when the difference between two consecutive terms decreases by the increase of n, as the terms' values tend to increase less and less until they converge at a fixed value.

The sum of the first terms of S₁(n) is

It is obvious that the difference between two consecutive terms increases more and more, as

and so on. Therefore, S₁(x) is divergent. Based on the above rule, so must be S₂(x) as well. Indeed,

It is obvious that the difference between two consecutive terms increases more and more, as

and so on. Therefore, S₂(x) is divergent.

Example 3

Check whether the following series are convergent or divergent by comparing them with a known series.

1. S₁ (n) = ∞∑_{n = 1}1/2 ∙ (3/4)ⁿ
2. S₂ (n) = ∞∑_{n = 1}7 ∙ 3ⁿ

Solution 3

1. We can write
   S₁ (n) = ∞∑_{n = 1}1/2 ∙ (3/4)ⁿ
   = 1/2 ∞∑_{n = 1}(3/4)ⁿ
   = 1/2 S(n)
   We can check the convergence of the known geometric series
   S(n) = ∞∑_{n = 1}(3/4)ⁿ
   to see whether S₁(n) is converging or not. We can write
   S(n) = ∞∑_{n = 1}(3/4)ⁿ
   = (3/4)¹ + (3/4)² + (3/4)³ + ⋯
   = (3/4) + (3/4) ∙ (3/4) + (3/4) ∙ (3/4)² + ⋯
   Thus, x₁ = 3/4 and R = 3/4. Therefore, since S(n) is an infinite series, we have
   S(n) = x₁/1 - R
   = 3/4/1 - 3/4
   = 3/4/4/4 - 3/4
   = 3/4/1/4
   = 3/4 ∙ 4/1
   = 12/4
   = 3
   Given that, this series is convergent, S₁(n) is convergent too, as
   S₁ (n) = 1/2 S(n)
   hence, it meets the first condition of the convergence rule.
2. We can write
   S₂ (n) = ∞∑_{n = 1}7 ∙ 3ⁿ
   = 7 ∙ ∞∑_{n = 1}3ⁿ
   = 7S(n)
   Let's check the convergence of the series
   S(n) = ∞∑_{n = 1}3ⁿ
   = 3¹ + 3² + 3³ + ⋯
   = 3 + 3 ∙ 3 + 3 ∙ 3² + ⋯
   = 3 + 9 + 27 + ⋯
   In this infinite geometric series, x₁ = 3 and R = 3. Hence, we cannot apply the convergence formula
   S(n) = x₁/1 - R
   because it is true only for series where -1 < R < 1. Indeed, it is evident that the terms increase more and more, as
   x₂ - x₁ = 9 - 3 = 6
   x₃ - x₂ = 27 - 9 = 18
   and so on. Therefore, the series S(n) is diverging. This means that since S₁(n) = 7Sn, then the original series S₁(n) is diverging too.

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