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?

Short answer

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

Step-by-step solution

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.

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 \(\int_{1}^{\infty} \frac{1}{x^{p}} dx\).

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: $$ \int_{1}^{\infty} \frac{1}{x^{p}} dx = \lim_{b \to \infty} \left[ \frac{x^{1-p}}{1 - p} \right]_1^b = \lim_{b \to \infty} \frac{b^{1-p}}{1 - p} - \frac{1^{1-p}}{1 - p} $$

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.

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 \(p \le 1\).