Question

Find a series that a. converges faster than \(\sum \frac{1}{k^{2}}\) but slower than \(\sum \frac{1}{k^{3}}\). b. Diverges faster than \(\sum \frac{1}{k}\) but slower than \(\sum \frac{1}{\sqrt{k}}\). c. Converges faster than \(\sum \frac{1}{k \ln ^{2} k}\) but slower than \(\sum \frac{1}{k^{2}}\).

Short answer

Question: Find a series that converges faster than the series \(\sum \frac{1}{k^{2}}\) but slower than the series \(\sum \frac{1}{k^{3}}\). Answer: The series \(\sum \frac{1}{k^{\frac{5}{2}}}\) converges faster than the series \(\sum \frac{1}{k^{2}}\) but slower than the series \(\sum \frac{1}{k^{3}}\).

Step-by-step solution

To find a series between the given series, we can choose an exponent between 2 and 3. Let's choose an exponent of \(\frac{5}{2}\). The series, then, becomes: \(\sum \frac{1}{k^{\frac{5}{2}}}\). #Step 2: Confirm the convergence of the series

To confirm that our chosen series converges, we can apply the integral test: \(\int_{1}^{\infty} \frac{1}{x^{\frac{5}{2}}} dx\). Since the exponent \(\frac{5}{2}>1\), the integral converges. Hence, the series \(\sum \frac{1}{k^{\frac{5}{2}}}\) converges faster than \(\sum \frac{1}{k^{2}}\) but slower than \(\sum \frac{1}{k^{3}}\) as required. #exercise#b: Find a series that diverges faster than \(\sum \frac{1}{k}\) but slower than \(\sum \frac{1}{\sqrt{k}}\). #Step 1: Deetermine a series diverging between the given series

To find a series between the given series, we can choose an exponent between 1 and \(1/2\). Let's choose an exponent of \(\frac{3}{4}\). The series, then, becomes: \(\sum \frac{1}{k^{\frac{3}{4}}}\). #Step 2: Confirm the divergence of the series

To confirm that our chosen series diverges, we can again apply the integral test: \(\int_{1}^{\infty} \frac{1}{x^{\frac{3}{4}}} dx\). Since the exponent \(\frac{3}{4}<1\), the integral diverges. Hence, the series \(\sum \frac{1}{k^{\frac{3}{4}}}\) diverges faster than \(\sum \frac{1}{k}\) but slower than \(\sum \frac{1}{\sqrt{k}}\) as required. #exercise#c: Find a series that converges faster than \(\sum \frac{1}{k \ln ^{2} k}\) but slower than \(\sum \frac{1}{k^{2}}\). #Step 1: Determine a series converging between the given series

To find a series between the given series, we can introduce an additional logarithmic factor with an exponent between 1 and 2. Let's choose an exponent of 1.5. The series, then, becomes: \(\sum \frac{1}{k \ln ^{\frac{3}{2}} k}\). #Step 2: Confirm the convergence of the series

While we cannot apply the classical integral test directly to confirm the convergence of the chosen series, we can use a comparison test. Since \(\sum \frac{1}{k \ln ^{2} k}\) converges (by the integral test with substitution) and \(\frac{1}{k \ln ^{\frac{3}{2}} k} < \frac{1}{k \ln ^{2} k}\) for sufficiently large \(k\), our chosen series converges by the comparison test. Hence, the series \(\sum \frac{1}{k \ln ^{\frac{3}{2}} k}\) converges faster than \(\sum \frac{1}{k \ln ^{2} k}\) but slower than \(\sum \frac{1}{k^{2}}\) as required.

Key concepts

Integral Test

The integral test is an essential tool used to determine the convergence or divergence of infinite series. This test compares an infinite series with an improper integral to establish whether the series converges or not. Here's how it works:

First, consider a series \(\sum_{k=1}^{\infty} a_k\) and a corresponding function \(f(x)\) that is continuous, positive, and decreasing for \(x \ge 1\). We then evaluate the improper integral \(\int_{1}^{\infty} f(x) \, dx\).

• If the integral converges, so does the series.
• If the integral diverges, the series diverges as well.

As an example, if we apply the integral test to \(\sum \frac{1}{k^{5/2}}\), we consider the integral \(\int_{1}^{\infty} \frac{1}{x^{5/2}} \, dx\). This integral converges, thus ensuring the convergence of the series.

Comparison Test

The comparison test is another powerful method for establishing the convergence or divergence of a series. It involves comparing a series with another series whose convergence properties are already known.

• To apply this test, consider two series \(\sum a_k\) and \(\sum b_k\). Suppose \(0 \le a_k \le b_k\) for all \(k\) beyond a certain point.
• If \(\sum b_k\) converges, then \(\sum a_k\) also converges.
• If \(\sum a_k\) diverges, then \(\sum b_k\) also diverges.

For instance, in exercise c, we compare \(\sum \frac{1}{k \ln^{3/2} k}\) with \(\sum \frac{1}{k \ln^{2} k}\), where the latter series is known to converge. Because \(\frac{1}{k \ln^{3/2} k} < \frac{1}{k \ln^{2} k}\) for sufficiently large \(k\), we can conclude that our chosen series converges using the comparison test.

P-Series

A p-series is a specific type of series that is expressed in the form \(\sum \frac{1}{k^p}\). The value of \(p\) determines whether the series converges or diverges:

• If \(p > 1\), the p-series converges.
• If \(p \le 1\), the p-series diverges.

The p-series serves as a benchmark for comparing the convergence of other series. In the original exercise, we see examples of this. When we assessed \(\sum \frac{1}{k^{5/2}}\) in comparison to \(\sum \frac{1}{k^{2}}\) and \(\sum \frac{1}{k^{3}}\), we used the known convergence properties of p-series to accurately determine how quickly the series converged.

Logarithmic Series

Logarithmic series involve terms with logarithmic expressions, often leading to more complex convergence analysis. A classic example is \(\sum \frac{1}{k \ln^{p} k}\). Similar to p-series, you can determine convergence based on the exponent \(p\).

• For \(p > 1\), the series converges.
• For \(p \le 1\), the series diverges.

In the step-by-step solution, we needed a series between \(\sum \frac{1}{k \ln^{2} k}\) and \(\sum \frac{1}{k^{2}}\). By choosing \(\sum \frac{1}{k \ln^{3/2} k}\), we adjusted the exponent to achieve convergence with the aid of the comparison test. This shows the flexibility of the logarithmic series to model scenarios requiring precise convergence tuning.