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Chapter 9: Problem 41

Integral Test \(\quad\) State the Integral Test and give an example of its use.

Short Answer

Expert verified
The Integral Test is used to determine the convergence or divergence of a series. It requires that the terms of the series can be expressed by a function that is continuous, positive, and decreasing on an interval. One example of its use is the series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\), where the function \(f(x) = \frac{1}{x^2}\) is used. Analysis via the Integral Test shows that the series converges.

Step by step solution

01

The Integral Test

The Integral Test is introduced in calculus to test the convergence or divergence of a series. The test states that if \(f\) is a continuous, positive, and decreasing function on \( [1, \, \infty) \) and \(a_n = f(n)\) for every integer \(n \geq 1\), then the series \(\sum_{n=1}^{\infty} a_n\) and the integral \(\int_{1}^{\infty} f(x) \, dx\) either both converge or both diverge.
02

Providing an Example

An example of using the Integral Test can be to determine whether the series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) converges or diverges. Consider the function \(f(x) = \frac{1}{x^2}\) on the interval \([1, \, \infty)\). This function is clearly positive, continuous and decreasing. The integral, \(\int_{1}^{\infty} f(x) \, dx = \int_{1}^{\infty} \frac{1}{x^2} \, dx\), evaluates to -1 at the upper limit and 1 at the lower limit, resulting in \(1 - (-1) = 2\). Since the integral is finite, according to the Integral Test, the series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) converges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convergence
When we talk about convergence in mathematics, we're exploring if a series adds up to a specific number. Imagine you're repeatedly adding smaller and smaller numbers.
Eventually, you might reach a point where adding more hardly changes the total. This behavior is what we call convergence.

For the Integral Test, if the integral of a function is finite, then the series converges.
  • This means you can find a single number it approaches.
  • Convergence means stability in the sum of infinite terms.
In our example, \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges because we achieved a finite result from the integral \( \int_{1}^{\infty} \frac{1}{x^2} \, dx \). This means as we keep adding terms like \( \frac{1}{1^2}, \frac{1}{2^2}, \dots \), the total will point to a definite number.
Series
A series is essentially a sum of a sequence, where you're adding numbers one after the other. You may have encountered series when dealing with patterns or sequences.
Mathematicians often explore if these sums have limits (converges) or not (diverges).

When employing the Integral Test, we pair up a series with an integral to understand its behavior.
  • The function \( f(x) \) must match the terms in the series, like \( a_n = f(n) \).
  • The series and corresponding function share characteristics like being positive, continuous, and decreasing.
This connection is powerful, as it allows us to determine the overall sum's nature by integrating a similar function.
Calculus
In calculus, you study change and motion through derivatives and integrals.
The Integral Test is a technique from calculus used to analyze series, leveraging the power of integration.

Here鈥檚 how it works in practice:
  • Identify the function \( f(x) \) that mirrors your series.
  • Ensure\( f(x) \) is positive, continuous, and decreases in value over time.
  • Compute the improper integral \( \int_{1}^{\infty} f(x) \, dx \).
If this integral is finite, the series converges; otherwise, it diverges. Calculus allows these steps to transform a series problem into an integration problem, which can often be easier to solve.
Divergence
Divergence is the opposite of convergence; it indicates that a series does not settle at a particular value.
If the addition of terms keeps increasing indefinitely, the series diverges.

With the Integral Test, evaluating an integral as infinite hints at divergence.
  • If you find \( \int f(x) \, dx \) growing without limit, the series diverges.
This could mean an infinite sum or even an undefined result. Understanding divergence helps recognize when a series doesn't stabilize, guiding you in approaching mathematical problems differently. The Integral Test thus becomes a valuable tool to detect these cases.

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Most popular questions from this chapter

Using a Binomial Series In Exercises \(17-26,\) use the binomial series to find the Maclaurin series for the function. $$ f(x)=\sqrt{1+x^{3}} $$

the terms of a series \(\sum_{n=1}^{\infty} a_{n}\) are defined recursively. Determine the convergence or divergence of the series. Explain your reasoning. \(a_{1}=\frac{1}{3}, a_{n+1}=\left(1+\frac{1}{n}\right) a_{n}\)

Approximating an Integral In Exercises \(63-70\) , use a power series to approximate the value of the integral with an error of less than \(0.0001 .\) (In Exercises 65 and \(67,\) assume that the integrand is defined as 1 when \(x=0 .\) $$ \int_{0}^{1 / 4} x \ln (x+1) d x $$

Approximating an Integral In Exercises \(63-70\) , use a power series to approximate the value of the integral with an error of less than \(0.0001 .\) (In Exercises 65 and \(67,\) assume that the integrand is defined as 1 when \(x=0 .\) $$ \int_{0}^{1} \frac{\sin x}{x} d x $$

Investigation Consider the function \(f\) defined by $$f(x)=\left\\{\begin{array}{ll}{e^{-1 / x^{2}},} & {x \neq 0} \\ {0,} & {x=0}\end{array}\right.$$ (a) Sketch a graph of the function. (b) Use the alternative form of the definition of the derivative (Section 2.1) and L'H贸pital's Rule to show that \(f^{\prime}(0)=0\) . [By continuing this process, it can be shown that \(f^{(n)}(0)=0\) for \(n>1 . ]\) (c) Using the result in part (b), find the Maclaurin series for \(f\) . Does the series converge to \(f ?\)
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