Integral Test

Integral Test is one of the simplest methods in calculus taught in terms of proving convergence or divergence in a given infinite series. It exhibits a connection between a series and an improper integral. By comparing a series to the integral of its terms, one can draw certain conclusions about the series. This test is particularly useful when dealing with series that involve functions that are difficult to analyze directly.

In this article, we will learn in detail about integral test, condition for integral test, its application and solved examples based on it.

What is the Integral Test?

The Integral Test is a test used in calculus to assess the convergence or divergence of an infinite series given in terms of the comparison with an improper integral. It states that if a series \sum_{n=1}^{\infty} a_n is positive, decreasing, and continuous, then the series converges if and only if the corresponding improper integral \int_{1}^{\infty} f(x) \, dx converges. This test is important because it is one method through which one can determine the convergence of some series whose terms do not follow a straightforward pattern or formula for their terms.

Integral Test for Convergence

The Integral Test is used to determine the convergence of an infinite series by comparing it to an improper integral. Given a series \sum_{n=1}^{\infty} a_n with positive terms, if the function f(x) = a_n is positive, decreasing, and continuous for all x \geq 1 then the series converges if and only if the integral \int_{1}^{\infty} f(x) \, dx converges. The derivation involves establishing the relationship between the series and the integral, ensuring the conditions for the Integral Test are met, and then evaluating the integral to determine convergence.

Conditions for Integral Test

The conditions for the Integral Test are as follows:

• The series must have positive terms: a_n > 0 for all n.
• The terms of the series must be decreasing: a_n+1 ≤ a_n for all n.
• The function f(x) = a_n must be continuous for all x ≥ 1

To conclude the convergence of the series \sum_{n=1}^{\infty} a_n use the formula:

\sum_{n=1}^{\infty} a_n converges if and only if \int_{1}^{\infty} f(x) \, dx converges

Integral Test Proof

The proof of the Integral Test concerns proving that a given series converges to the one stated in the theorem, depending on the conditions associated with the function. Let's go through the key steps:

1. Show that the terms of the series are positive and decreasing:

• Let a_n = f(n), where f(x) is a positive, decreasing, and continuous function for x ≥ 1
• Since f(x) is decreasing, we have f(n+1) ≤ f(n) for all n ≥ 1, which implies a_n+1 ≤ a_n.
• Therefore, the terms of the series are positive and decreasing.

2. Establish the relationship between the series and the integral:

• Consider the partial sums of the series: S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} f(k)
• We can bound the partial sums using integrals: \int_{1}^{n+1} f(x) \, dx \geq \sum_{k=1}^{n} f(k) \geq \int_{1}^{n} f(x) \, dx
• This is because each term f(k) is greater than or equal to the area of the rectangle with base [k−1, k] and height f(k), but less than or equal to the area of the rectangle with base [k−1, k] and height f(k−1).

3. Evaluate the integral to determine convergence:

• If the improper integral \int_{1}^{\infty} f(x) \, dx converges, then the sequence of partial integrals \int_{1}^{n} f(x) \, dx converges as n→∞.
• Since \int_{1}^{n} f(x) \, dx \geq \sum_{k=1}^{n-1} f(k) = S_{n-1}, the sequence of partial sums ?n also converges.
• Conversely, if the series \sum_{n=1}^{\infty} a_n converges, then the sequence of partial sums ?n converges, and hence the sequence of partial integrals \int_{1}^{n} f(x) \, dx also converges.
• By the properties of improper integrals, this implies that the improper integral \int_{1}^{\infty} f(x) \, dx converges.

Therefore, we have shown that the series \sum_{n=1}^{\infty} a_n converges if and only if the improper integral \int_{1}^{\infty} f(x) \, dx converges, provided that the terms of the series are positive and decreasing, and the function f(x) is continuous for x ≥ 1. This completes the proof of the Integral Test.

Comparison Test

Comparison test is used in the determination of the convergence of a series through comparison to its co-efficient series whose convergence is known. The thought process is to look for a series whose status of convergence is either proven or can be used to compare with the given series.

Let \sum_{n=1}^{\infty} a_n and \sum_{n=1}^{\infty} b_n be two series with non-negative terms. If there exists a constant ? > 0 such that 0 ≤ ?n ≤ ?⋅?n for all n ≥ 1 then:

• If \sum_{n=1}^{\infty} b_n converges, then \sum_{n=1}^{\infty} a_n also converges
• If \sum_{n=1}^{\infty} a_n converges, then \sum_{n=1}^{\infty} b_n also converges

Applications of Integral Test

The applications of integral tests are as follows:

• Determining convergence of series with non-elementary terms.
• Analyzing the convergence of alternating series.
• Estimating the sum of series with known integrals.
• Investigating the convergence of power series.
• Studying the behavior of series involving trigonometric functions.

Solved Examples on Integral Test

Example 1: Determine the convergence of the series \sum_{n=1}^{\infty} \frac{1}{n^p} using the Integral Test.

Solution:

Let f(x) = \frac{1}{x^p}. We want to compare the series \sum_{n=1}^{\infty} \frac{1}{n^p} to the improper integral \int_{1}^{\infty} \frac{1}{x^p} \, dx.

The function f(x) = \frac{1}{x^p} is positive, decreasing, and continuous for x \geq 1 and all p > 0.

Evaluating the integral:

\int_{1}^{\infty} \frac{1}{x^p} \, dx = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^p} \, dx

= \lim_{b \to \infty} \left[ \frac{-1}{p-1} x^{-(p-1)} \right]_{1}^{b}

= \lim_{b \to \infty} \left( \frac{-1}{p-1} b^{-(p-1)} + \frac{1}{p-1} \right)

If p > 1, then the integral converges, and by the Integral Test, the series \sum_{n=1}^{\infty} \frac{1}{n^p} also converges.

If p \leq 1, then the integral diverges, and by the Integral Test, the series \sum_{n=1}^{\infty} \frac{1}{n^p} also diverges.

Therefore, the series \sum_{n=1}^{\infty} \frac{1}{n^p} converges if p > 1 and diverges if p \leq 1.

Example 2: Determine the convergence of the series \sum_{n=1}^{\infty} \frac{1}{n \ln n} using the Integral Test.

Solution:

Let f(x) = \frac{1}{x \ln x}. We want to compare the series \sum_{n=1}^{\infty} \frac{1}{n \ln n} to the improper integral \int_{1}^{\infty} \frac{1}{x \ln x} \, dx.

The function f(x) = \frac{1}{x \ln x} is positive, decreasing, and continuous for x \geq 1.

Evaluating the integral:

\int_{1}^{\infty} \frac{1}{x \ln x} \, dx = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x \ln x} \, dx

= \lim_{b \to \infty} \left[ \ln \ln x \right]_{1}^{b}

= \lim_{b \to \infty} \left( \ln \ln b - \ln \ln 1 \right)

= \lim_{b \to \infty} \ln \ln b

Since \lim_{b \to \infty} \ln \ln b = \infty, the integral diverges, and by the Integral Test, the series \sum_{n=1}^{\infty} \frac{1}{n \ln n} also diverges.

Therefore, the series \sum_{n=1}^{\infty} \frac{1}{n \ln n} diverges.

Example 3: Determine the convergence of the series \sum_{n=1}^{\infty} \frac{1}{n^2 + 1} using the Integral Test.

Solution:

Let f(x) = \frac{1}{x^2 + 1}. We want to compare the series \sum_{n=1}^{\infty} \frac{1}{n^2 + 1} to the improper integral \int_{1}^{\infty} \frac{1}{x^2 + 1} \, dx.

The function f(x) = \frac{1}{x^2 + 1} is positive, decreasing, and continuous for x \geq 1.

Evaluating the integral:

\int_{1}^{\infty} \frac{1}{x^2 + 1} \, dx = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^2 + 1} \, dx

= \lim_{b \to \infty} \left[ \arctan x \right]_{1}^{b}

= \lim_{b \to \infty} \left( \arctan b - \arctan 1 \right)

= \lim_{b \to \infty} \arctan b - \arctan 1

= \frac{\pi}{2} - \frac{\pi}{4}

= \frac{\pi}{4}

Since the integral converges, by the Integral Test, the series \sum_{n=1}^{\infty} \frac{1}{n^2 + 1} also converges.

Therefore, the series \sum_{n=1}^{\infty} \frac{1}{n^2 + 1} converges.

Practice Questions on the Integral Test

Q1: Determine the convergence or divergence of the series \sum_{n=1}^{\infty} \frac{1}{n^2 + n} using the Integral Test.

Q2: Determine the convergence or divergence of the series \sum_{n=1}^{\infty} \frac{1}{n^2 + 2n + 1} using the Integral Test.

Q3: Determine the convergence or divergence of the series \sum_{n=1}^{\infty} \frac{1}{n^3 + 1} using the Integral Test.

Conclusion

In conclusion, the Integral Test is a valuable tool in calculus for analyzing the convergence of infinite series by relating them to improper integrals. By establishing conditions for convergence, comparing series to integrals, and rigorously proving convergence or divergence, this test provides a systematic approach to understanding the behavior of series with complex terms. Its applications extend to various mathematical contexts, making it a fundamental technique in the study of infinite series.

Also, Check

• Integration in Math
• Integrals
• Integration Formulas