Problem 4 - Calculus Early Transcendentals

Question

For what values of \(p\) does the series \(\sum_{k=1}^{\infty} \frac{1}{k^{p}}\) converge? For what values of \(p\) does it diverge?

Step-by-Step Solution

Answer

Answer: The series converges for p > 1 and diverges for p ≤ 1.

Step 1: Express the series with the general formula

We are given the series \(\sum_{k=1}^{\infty} \frac{1}{k^{p}}\). Notice that the series starts at \(k = 1\) and goes to infinity.

Step 2: Apply the Integral Test

Now, we will use the integral test to determine the convergence of the series. We need to compare the given series with the improper integral \(\int_{1}^{\infty} \frac{1}{x^{p}} dx\).

Step 3: Evaluate the improper integral

First, we'll find the antiderivative of the function \(\frac{1}{x^{p}}\). The antiderivative is given by: $$ \int \frac{1}{x^{p}} dx = \frac{x^{1-p}}{1 - p} + C $$ where C is the constant of integration. Now, we'll evaluate the improper integral: $$ \int_{1}^{\infty} \frac{1}{x^{p}} dx = \lim_{b \to \infty} \left[ \frac{x^{1-p}}{1 - p} \right]_1^b = \lim_{b \to \infty} \frac{b^{1-p}}{1 - p} - \frac{1^{1-p}}{1 - p} $$

Step 4: Determine the convergence of the improper integral

Now, we'll examine the limit of the improper integral as \(b\) goes to infinity: $$ \lim_{b \to \infty} \frac{b^{1-p}}{1 - p} - \frac{1^{1-p}}{1 - p} $$ We can now see that if \(p > 1\), the term \(b^{1-p}\) approaches 0 as \(b\) goes to infinity, and the integral converges. However, if \(p \le 1\), the term \(b^{1-p}\) either goes to infinity or stays constant, and the integral diverges.

Step 5: Conclusion

Using the integral test, we find that the series converges for p > 1 and diverges for p ≤ 1. Therefore, the series \(\sum_{k=1}^{\infty} \frac{1}{k^{p}}\) converges for \(p > 1\) and diverges for \(p \le 1\).