What Is The Limit Comparison Test

The Limit Comparison Test is a powerful tool in calculus used to determine the convergence or divergence of an infinite series by comparing it to another series whose convergence behavior is already known. It's particularly useful when dealing with series that resemble p-series or geometric series. This test provides a straightforward way to ascertain the behavior of a series without needing to find its exact sum.

Understanding the Basics

Before diving into the specifics of the Limit Comparison Test, let's recap some fundamental concepts about infinite series and convergence. An infinite series is the sum of an infinite sequence of numbers. The series converges if the sequence of its partial sums approaches a finite limit; otherwise, it diverges. Determining whether a series converges or diverges is crucial in many areas of mathematics, physics, and engineering.

Convergence vs. Divergence

• Convergence: A series ∑an converges if the sequence of its partial sums, Sn = a1 + a2 + ... + an, approaches a finite limit as n approaches infinity. In other words, there exists a number L such that lim(n→∞) Sn = L.
• Divergence: A series ∑an diverges if the sequence of its partial sums does not approach a finite limit as n approaches infinity. This can happen if the partial sums grow without bound or oscillate indefinitely.

Why Use Comparison Tests?

Directly determining the convergence or divergence of a series can be challenging. Comparison tests offer an alternative by relating the given series to another series with known convergence properties. The Limit Comparison Test is particularly advantageous when the Direct Comparison Test is difficult to apply.

The Limit Comparison Test: A Detailed Explanation

The Limit Comparison Test states that for two series ∑an and ∑bn with positive terms, if the limit as n approaches infinity of the ratio an/bn exists and is a finite positive number, then both series either converge or diverge together.

Formal Statement

Let ∑an and ∑bn be series with positive terms. If

lim (n→∞) an/bn = c

where 0 < c < ∞, then either both series converge or both series diverge.

Conditions for the Test

1. Positive Terms: Both series ∑an and ∑bn must have positive terms for all n. This condition is essential because the test relies on the ratio an/bn being positive, which simplifies the analysis.
2. Limit Exists and is Positive: The limit of the ratio an/bn as n approaches infinity must exist and be a finite positive number. If the limit is zero, infinity, or does not exist, the test is inconclusive, and another method must be used.

How to Apply the Limit Comparison Test

1. Choose a Comparison Series: Select a series ∑bn whose convergence behavior is known. This series should be similar to the given series ∑an in some way. Common choices include p-series (∑1/np) and geometric series (∑arn-1).

2. Compute the Limit: Calculate the limit of the ratio an/bn as n approaches infinity:

   lim (n→∞) an/bn

3. Evaluate the Limit:

   • If the limit is a finite positive number (0 < c < ∞), then ∑an and ∑bn either both converge or both diverge.
   • If the limit is 0 or ∞, the test is inconclusive, and you must use a different test.

4. Determine Convergence: Based on the convergence behavior of ∑bn, conclude whether ∑an converges or diverges.

Example 1: Convergent Series

Consider the series:

∑ (n=1 to ∞) (3n - 2) / (n^3 - n + 1)

We want to determine whether this series converges or diverges.

1. Choose a Comparison Series: The dominant terms in the numerator and denominator are 3n and n^3, respectively. Therefore, we can compare it with the series ∑bn = ∑ 3n / n^3 = ∑ 3 / n^2. Since ∑ 1 / n^2 is a convergent p-series (p = 2 > 1), ∑ 3 / n^2 also converges.

2. Compute the Limit: Calculate the limit of the ratio an/bn as n approaches infinity:

   lim (n→∞) [(3n - 2) / (n^3 - n + 1)] / [3 / n^2]

   = lim (n→∞) [(3n - 2) / (n^3 - n + 1)] * [n^2 / 3]

   = lim (n→∞) (3n^3 - 2n^2) / (3n^3 - 3n + 3)

3. Evaluate the Limit: Divide both the numerator and denominator by n^3:

   lim (n→∞) (3 - 2/n) / (3 - 3/n^2 + 3/n^3) = 3 / 3 = 1

4. Determine Convergence: Since the limit is 1, which is a finite positive number, and ∑ 3 / n^2 converges, the given series ∑ (3n - 2) / (n^3 - n + 1) also converges.

Example 2: Divergent Series

Consider the series:

∑ (n=1 to ∞) (2n^2 + 3n) / (n^3 + 1)

We want to determine whether this series converges or diverges.

1. Choose a Comparison Series: The dominant terms in the numerator and denominator are 2n^2 and n^3, respectively. Therefore, we can compare it with the series ∑bn = ∑ 2n^2 / n^3 = ∑ 2 / n. Since ∑ 1 / n is the harmonic series, which is a divergent p-series (p = 1), ∑ 2 / n also diverges.

2. Compute the Limit: Calculate the limit of the ratio an/bn as n approaches infinity:

   lim (n→∞) [(2n^2 + 3n) / (n^3 + 1)] / [2 / n]

   = lim (n→∞) [(2n^2 + 3n) / (n^3 + 1)] * [n / 2]

   = lim (n→∞) (2n^3 + 3n^2) / (2n^3 + 2)

3. Evaluate the Limit: Divide both the numerator and denominator by n^3:

   lim (n→∞) (2 + 3/n) / (2 + 2/n^3) = 2 / 2 = 1

4. Determine Convergence: Since the limit is 1, which is a finite positive number, and ∑ 2 / n diverges, the given series ∑ (2n^2 + 3n) / (n^3 + 1) also diverges.

Tips for Choosing the Comparison Series

Choosing the right comparison series is crucial for successfully applying the Limit Comparison Test. Here are some tips to guide your choice:

1. Identify Dominant Terms: Look for the dominant terms in the numerator and denominator of the series' terms. These terms often dictate the series' overall behavior.
2. Consider p-series: p-series (∑1/np) are excellent choices when the terms of the series involve powers of n. Remember that a p-series converges if p > 1 and diverges if p ≤ 1.
3. Consider Geometric Series: Geometric series (∑arn-1) are useful when the terms of the series involve powers of a constant. A geometric series converges if |r| < 1 and diverges if |r| ≥ 1.
4. Simplify Complex Expressions: Simplify complex expressions to identify the essential components that determine the series' behavior.
5. Practice: The more you practice, the better you'll become at recognizing suitable comparison series.

When the Limit Comparison Test is Inconclusive

The Limit Comparison Test is not always conclusive. If the limit of the ratio an/bn as n approaches infinity is 0 or ∞, the test provides no information about the convergence or divergence of the series ∑an. In such cases, you must use a different test, such as the Direct Comparison Test, the Ratio Test, the Root Test, or the Integral Test.

Example 3: Inconclusive Case (Limit is Zero)

Consider the series:

∑ (n=1 to ∞) 1 / (n^3 + n)

Suppose we try to compare this series with the harmonic series ∑bn = ∑ 1 / n, which is known to diverge.

1. Compute the Limit: Calculate the limit of the ratio an/bn as n approaches infinity:

   lim (n→∞) [1 / (n^3 + n)] / [1 / n]

   = lim (n→∞) [1 / (n^3 + n)] * [n / 1]

   = lim (n→∞) n / (n^3 + n)

2. Evaluate the Limit: Divide both the numerator and denominator by n:

   lim (n→∞) 1 / (n^2 + 1) = 0

3. Determine Convergence: Since the limit is 0, the Limit Comparison Test is inconclusive.

However, we can use the Direct Comparison Test. Since n^3 + n > n^3 for all n ≥ 1, we have 1 / (n^3 + n) < 1 / n^3. The series ∑ 1 / n^3 is a convergent p-series (p = 3 > 1), so by the Direct Comparison Test, the series ∑ 1 / (n^3 + n) also converges.

Example 4: Inconclusive Case (Limit is Infinity)

Consider the series:

∑ (n=1 to ∞) (n^2 + 1) / n

Suppose we try to compare this series with the convergent p-series ∑bn = ∑ 1 / n^2.

1. Compute the Limit: Calculate the limit of the ratio an/bn as n approaches infinity:

   lim (n→∞) [(n^2 + 1) / n] / [1 / n^2]

   = lim (n→∞) [(n^2 + 1) / n] * [n^2 / 1]

   = lim (n→∞) (n^4 + n^2) / n

2. Evaluate the Limit: Divide both the numerator and denominator by n:

   lim (n→∞) n^3 + n = ∞

3. Determine Convergence: Since the limit is ∞, the Limit Comparison Test is inconclusive.

However, we can observe that the terms (n^2 + 1) / n = n + 1/n do not approach zero as n approaches infinity. Therefore, by the Divergence Test (also known as the nth-Term Test), the series ∑ (n^2 + 1) / n diverges.

Advantages and Disadvantages

Advantages

• Simplicity: The Limit Comparison Test is often easier to apply than the Direct Comparison Test, especially when dealing with complex expressions.
• Flexibility: It allows for a broader range of comparison series to be used, increasing the chances of finding a suitable one.
• Intuitive: The test's rationale is straightforward: if two series have terms that are asymptotically proportional, they should have the same convergence behavior.

Disadvantages

• Requires Positive Terms: The test is only applicable to series with positive terms.
• Inconclusive Cases: The test can be inconclusive if the limit of the ratio is 0 or ∞, requiring the use of other tests.
• Choice of Comparison Series: The success of the test depends on choosing an appropriate comparison series, which may not always be obvious.

Common Mistakes to Avoid

1. Forgetting to Check for Positive Terms: Ensure that both series have positive terms before applying the test.
2. Incorrectly Calculating the Limit: Double-check your calculations when finding the limit of the ratio an/bn.
3. Misinterpreting Inconclusive Results: Understand that a limit of 0 or ∞ does not provide any information about the convergence or divergence of the series.
4. Choosing an Unsuitable Comparison Series: Select a comparison series that is similar to the given series and whose convergence behavior is known.
5. Ignoring the Importance of Practice: Practice applying the test to various examples to improve your skills and intuition.

Applications of the Limit Comparison Test

The Limit Comparison Test is a valuable tool in various areas of mathematics and its applications. Here are some examples:

1. Determining the Convergence of Improper Integrals: The convergence of certain improper integrals can be related to the convergence of infinite series, and the Limit Comparison Test can be used to analyze these series.
2. Analyzing Power Series: The convergence of power series is crucial in complex analysis and differential equations. The Limit Comparison Test can help determine the radius of convergence of a power series.
3. Approximating Functions: Infinite series are often used to approximate functions. The Limit Comparison Test can help ensure that the approximation is valid by verifying the convergence of the series.
4. Physics and Engineering: Many physical phenomena are modeled using infinite series. The Limit Comparison Test can be used to analyze the stability and behavior of these models.

Examples With Detailed Steps

Example 5

Determine whether the series ∑ (n=1 to ∞) (√(n) + 1) / (n^2 + 2) converges or diverges.

1. Choose a Comparison Series: The dominant term in the numerator is √(n) = n^(1/2), and in the denominator, it is n^2. Thus, we compare with bn = n^(1/2) / n^2 = 1 / n^(3/2). The series ∑ 1 / n^(3/2) is a p-series with p = 3/2 > 1, so it converges.
2. Compute the Limit: lim (n→∞) [ (√(n) + 1) / (n^2 + 2) ] / [ 1 / n^(3/2) ] = lim (n→∞) [ (√(n) + 1) / (n^2 + 2) ] * [ n^(3/2) / 1 ] = lim (n→∞) (n^2 + n^(3/2)) / (n^2 + 2)
3. Evaluate the Limit: Divide both the numerator and denominator by n^2: lim (n→∞) (1 + 1/√n) / (1 + 2/n^2) = (1 + 0) / (1 + 0) = 1
4. Determine Convergence: Since the limit is 1 (a finite positive number) and ∑ 1 / n^(3/2) converges, the series ∑ (√(n) + 1) / (n^2 + 2) also converges.

Example 6

Determine whether the series ∑ (n=1 to ∞) (5n^3 - 2n) / (n^4 + 1) converges or diverges.

1. Choose a Comparison Series: The dominant term in the numerator is 5n^3, and in the denominator, it is n^4. Thus, we compare with bn = 5n^3 / n^4 = 5 / n. The series ∑ 5 / n is a harmonic series (∑ 1/n) multiplied by a constant, so it diverges.
2. Compute the Limit: lim (n→∞) [ (5n^3 - 2n) / (n^4 + 1) ] / [ 5 / n ] = lim (n→∞) [ (5n^3 - 2n) / (n^4 + 1) ] * [ n / 5 ] = lim (n→∞) (5n^4 - 2n^2) / (5n^4 + 5)
3. Evaluate the Limit: Divide both the numerator and denominator by n^4: lim (n→∞) (5 - 2/n^2) / (5 + 5/n^4) = (5 - 0) / (5 + 0) = 1
4. Determine Convergence: Since the limit is 1 (a finite positive number) and ∑ 5 / n diverges, the series ∑ (5n^3 - 2n) / (n^4 + 1) also diverges.

Example 7

Determine whether the series ∑ (n=1 to ∞) (4^n + n^2) / (n!) converges or diverges.

1. Choose a Comparison Series: In this case, it is difficult to directly compare with a p-series or a geometric series. However, we can still use the Limit Comparison Test with a simpler series involving factorials. Let's compare with bn = 4^n / n!.
2. Compute the Limit: lim (n→∞) [ (4^n + n^2) / (n!) ] / [ 4^n / (n!) ] = lim (n→∞) [ (4^n + n^2) / (n!) ] * [ (n!) / 4^n ] = lim (n→∞) (4^n + n^2) / 4^n
3. Evaluate the Limit: Divide both terms in the numerator by 4^n: lim (n→∞) (1 + n^2 / 4^n) Since exponential functions grow faster than polynomial functions, lim (n→∞) n^2 / 4^n = 0. Therefore, lim (n→∞) (1 + n^2 / 4^n) = 1 + 0 = 1
4. Determine Convergence: The series ∑ 4^n / n! converges (this can be shown using the Ratio Test). Since the limit is 1 (a finite positive number) and ∑ 4^n / n! converges, the series ∑ (4^n + n^2) / (n!) also converges.

Conclusion

The Limit Comparison Test is a versatile and valuable tool for determining the convergence or divergence of infinite series. By comparing a given series with another series whose convergence behavior is known, this test provides a straightforward way to analyze the series. While it has its limitations, understanding how to apply the Limit Comparison Test effectively can greatly simplify the process of determining the convergence of various series. Remember to always check the conditions of the test, choose an appropriate comparison series, and practice applying the test to various examples to improve your skills.