Question

Give an argument similar to that given in the text for the harmonic series to show that \(\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}\) diverges.

Short answer

Answer: Yes, the series \(\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}\) diverges.

Step-by-step solution

Define the series and the function to be compared.

Let's consider the series \(\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}\) and the function \(f(x) = \frac{1}{\sqrt{x}}\).

Apply the comparison test using the integral.

We will now use the comparison test to investigate the convergence of the series. To do this, we will compare the sum of the series with the integral of the function \(f(x)\) from 1 to infinity: $$\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}} \geq \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx$$

Evaluate the integral.

To evaluate the integral, we can compute: $$\begin{aligned} \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx &= \int_{1}^{\infty} x^{-\frac{1}{2}}\, dx \\ &= \left[\frac{x^{1-\frac{1}{2}}}{1-\frac{1}{2}} \right]_{1}^{\infty} \\ &= 2[\sqrt{x}]_{1}^{\infty} \end{aligned}$$

Analyze the result of the integral.

As \(x\) goes to infinity, \(\sqrt{x}\) also goes to infinity, so the evaluated integral becomes: $$2[\sqrt{x}]_{1}^{\infty} = 2(\infty - \sqrt{1}) = 2(\infty - 1) = \infty$$

Conclude.

Since the integral evaluates to infinity, we have: $$\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}} \geq \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx = \infty$$ This implies that the series \(\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}\) also diverges, similar to the harmonic series.

Key concepts

Comparison Test

The comparison test is a handy tool for determining the convergence or divergence of an infinite series. This test involves comparing the given series to another series whose behavior is already known. If you have a series \( \sum a_n \) and another series \( \sum b_n \), where eventually \( 0 \leq a_n \leq b_n \) for all \( n \) sufficiently large:

• If \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• If \( \sum a_n \) diverges, then \( \sum b_n \) also diverges.

To use the comparison test, ensure the terms of the series are non-negative. In our case, we compare the series \( \sum \frac{1}{\sqrt{k}} \) with an improper integral to analyze the series' behavior.

Integral Test

The integral test connects the sum of a series with an improper integral to determine convergence or divergence. If you have a series \( \sum_{k=1}^{\infty} a_k \) with positive, decreasing terms, consider the corresponding function \( f(x) \) such that \( a_k = f(k) \). The integral test states:

• If \( \int_{1}^{\infty} f(x) \, dx \) converges, then \( \sum_{k=1}^{\infty} a_k \) converges.
• If \( \int_{1}^{\infty} f(x) \, dx \) diverges, then \( \sum_{k=1}^{\infty} a_k \) diverges.

For \( \sum \frac{1}{\sqrt{k}} \), we set \( f(x) = \frac{1}{\sqrt{x}} \) and integrate from 1 to infinity. Since this improper integral diverges, it follows that our series also diverges.

Harmonic Series

A harmonic series is a specific infinite series that takes the form \( \sum_{k=1}^{\infty} \frac{1}{k} \). The harmonic series is well-known for its divergence, a property that might seem counterintuitive given the shrinking size of its terms.If you construct partial sums of the harmonic series, the series appears to grow without bound. This divergence relates closely to how quickly the terms decrease. Despite the terms approaching zero, their sum increases to infinity. Recognizing this behavior helps us understand similar series, such as \( \sum \frac{1}{\sqrt{k}} \), which also diverges due to similar slow-decaying terms.

Improper Integrals

Improper integrals extend the concept of integrals to functions with unbounded intervals or undefined points. When evaluating improper integrals, replace infinity with a limit to determine their behavior more accurately.For example, \( \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx \) involves taking the limit as the upper bound approaches infinity. In practical terms,

• Evaluate the antiderivative of the function in a finite interval.
• Use limits to assess appropriate behavior as you approach infinity.

Since this integral diverges, it implies the related series \( \sum \frac{1}{\sqrt{k}} \) also diverges by the integral test. Recognizing and working with improper integrals are crucial in understanding the behavior of series they are compared to.