Question

Use the integral test to prove the following so-called \(p\) -series test. The series $$\sum_{n=1}^{\infty} \frac{1}{n^{p}} \text { is }\left\\{\begin{array}{ll} \text { convergent } & \text { if } p > 1 \\ \text { divergent } & \text { if } p \leq 1 \end{array}\right.$$

Short answer

The series \( \sum_{n=1}^{\infty} \frac{1}{n^{p}} \) converges if \( p > 1 \) and diverges if \( p \leq 1 \).

Step-by-step solution

Set Up the Integral Test

To apply the integral test, consider the function \( f(x) = \frac{1}{x^p} \). We need to check if \( f(x) \) is positive, continuous, and decreasing for \( x \geq 1 \).

Verify Conditions

The function \( f(x) = \frac{1}{x^p} \) is clearly positive for all \( x > 0 \) as long as \( p \) is a positive number. It is continuous for all \( x > 0 \). Next, taking its derivative, \( f'(x) = -\frac{p}{x^{p+1}} \), which is negative for all \( x \geq 1 \) if \( p > 0 \), proving that \( f(x) \) is decreasing.

Set Up the Improper Integral

To apply the integral test, evaluate the improper integral \[ \int_1^{\infty} \frac{1}{x^p} \,dx. \]

Solve the Integral for \( p eq 1 \)

Evaluate the improper integral for \( p eq 1 \): \[ \int_1^{\infty} \frac{1}{x^p} \,dx = \lim_{b \to \infty} \int_1^b \frac{1}{x^p} \,dx \]. The antiderivative is \[ \int \frac{1}{x^p} \,dx = \frac{x^{1-p}}{1-p} + C \]. Thus, \[ \lim_{b \to \infty} \left[ \frac{b^{1-p}}{1-p} - \frac{1^{1-p}}{1-p} \right] \].

Analyze Convergence

We need to analyze the limit behavior of \( \lim_{b \to \infty} \frac{b^{1-p}}{1-p} \). - If \( p > 1 \), \( 1-p < 0 \) and \( \lim_{b \to \infty} b^{1-p} \to 0 \). The integral converges.- If \( p \leq 1 \), \( 1-p \geq 0 \) and \( \lim_{b \to \infty} b^{1-p} \to \infty \) or to a constant (for \( p=1 \), it diverges.

Conclusion

Therefore, the integral converges when \( p > 1 \), and diverges when \( p \leq 1 \). By the integral test, the series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) is convergent if \( p > 1 \) and divergent if \( p \leq 1 \).

Key concepts

integral test

The integral test is a powerful method used to determine the convergence or divergence of a mathematical series. It's particularly handy for series where the terms involve a function that is continuous, positive, and decreasing. The fundamental idea is to compare the series to an improper integral. If the improper integral converges, then so does the series, and if it diverges, so does the series.
To use the integral test, follow these steps:

• Formulate the corresponding function from the series terms.
• Ensure the function is continuous, positive, and decreases for all values greater than or equal to 1.
• Set up the improper integral from 1 to infinity.
• Solve the improper integral to determine its convergence or divergence.

Applying this test simplifies evaluating series by transforming the summation problem into an integration problem.

convergence

Convergence is a key concept when working with series. A series converges if the sum of its infinite terms approaches a finite number. Simply put, if you keep adding more terms of the series, and they settle down to a particular value, the series is convergent.
In the context of the integral test, we determine convergence by finding out if the improper integral of the related function is finite.

• If the improper integral \(\int_1^{\infty} f(x) \, dx\) converges, then the series \(\sum_{n=1}^{\infty} f(n)\) also converges.
• If it diverges, then the series also diverges.

Using our example, when \(\boldsymbol{p > 1}\), the function \(\frac{1}{x^p}\) results in a convergent integral, hence the series converges.

improper integral

An improper integral is an integral where the interval of integration is infinite or the integrand has an infinite discontinuity. These are used extensively in the integral test. Evaluating an improper integral involves careful handling of limits.
For instance, consider the integral of \(\frac{1}{x^p}\) from 1 to infinity:
\[ \int_1^{\infty} \frac{1}{x^p} \, dx \].
To solve this, we first compute the integral over a finite interval and then take the limit as the upper bound approaches infinity:
\[ \lim_{b \to \infty} \int_1^b \frac{1}{x^p} \, dx \].
The antiderivative of \(\frac{1}{x^p}\) is \(\frac{x^{1-p}}{1-p}\), and then we analyze the limit for different values of \p\. When \p > 1\, the integral converges and when \p \leq 1\, it diverges.

mathematical series

A mathematical series is the sum of the terms of a sequence. Series can be finite or infinite, and studying their convergence properties helps us understand their behavior.
In this exercise, we deal with the \(\boldsymbol{p}\)\-series, \(\sum\_{n=1}^{\infty} \frac{1}{n^p}\), where \p\ is any real number. The goal is to determine the criteria for convergence.
Series are essential in many areas of mathematics and physics, providing solutions to complex problems and allowing us to approximate functions using sums.

• A series converges if the sequence of partial sums converges to a limit.
• Divergence means the partial sums do not approach any limit.
• Using convergence tests like the integral test helps us understand and predict the behavior of infinite series.

This helps us apply these series in practical applications like Fourier series in signal processing and Taylor series in calculus.