Question

Find the values of the parameter \(p>0\) for which the following series converge. $$\sum_{k=1}^{\infty} \ln \left(\frac{k}{k+1}\right)^{p}$$

Short answer

Answer: The series converges for values of the parameter \(p > 1\).

Step-by-step solution

Simplify the expression

First, let's simplify the given series by rewriting the logarithm: $$\sum_{k=1}^{\infty} \ln \left(\frac{k}{k+1}\right)^{p} = \sum_{k=1}^{\infty} p \ln \left(\frac{k}{k+1}\right)$$

Create inequalities

Now, we need to find a series to compare it to. Notice that the function \(ln(\frac{k}{k + 1})\) can be compared with the function \(\frac{1}{k+1}\). This is because, as k increases, the natural logarithm becomes smaller, so: $$0 \leq p \ln \left(\frac{k}{k+1}\right) \leq p \frac{1}{k+1}$$

Apply the comparison test

Now, if we can determine whether the series \(\sum_{k=1}^{\infty} p \frac{1}{k+1}\) converges or diverges, then we can use the comparison test to determine the convergence of the original series. This new series can be written as \(p \sum_{k=1}^{\infty} \frac{1}{k+1}\). Since the series \(\sum_{k=1}^{\infty} \frac{1}{k+1}\) is a harmonic series with a translation, it converges if and only if the parameter \(p > 1\). Therefore, the series converges for any value of \(p > 1\).

State the final result

The original series $$\sum_{k=1}^{\infty} \ln \left(\frac{k}{k+1}\right)^{p}$$ converges for values of the parameter \(p > 1\).

Key concepts

Comparison Test

When we talk about series convergence, we often encounter situations where direct analysis of the series isn't straightforward. The Comparison Test is a valuable tool for such cases. This test helps us determine if a series converges or diverges by comparing it to another series that we already know the behavior of.

The idea is simple:

• If the series we are examining is smaller term-by-term than a known convergent series, then it also converges.
• Conversely, if it is larger than a known divergent series, it diverges.

In equation format, if we have two series, \(\sum a_k\) and \(\sum b_k\), then:

• If \(0 \leq a_k \leq b_k\) for all \(k\) and \(\sum b_k\) converges, then \(\sum a_k\) converges.
• If \(0 \leq b_k \leq a_k\) for all \(k\) and \(\sum b_k\) diverges, then \(\sum a_k\) diverges.

Applying this to our example, we compared \(p \ln\left(\frac{k}{k+1}\right)\) with \(p \frac{1}{k+1}\), knowing the behavior of the harmonic series to help us solve the problem.

Logarithmic Function

The logarithmic function, commonly denoted as \(\ln(x)\), is another core aspect of this problem. It is the inverse operation of exponentiation and has unique properties that make it useful in mathematical analysis.

One of the reasons we encounter logarithms in series analysis is because they provide a way to handle very gradual growth. Specifically, for large values of \(k\), \(\ln\left(\frac{k}{k+1}\right)\) simplifies neatly into a form that allows clean boundary comparison. This simplification plays a crucial role in applying the Comparison Test.

Remember, the function \(\ln(\frac{k}{k+1})\) tends to decrease as \(k\) increases. This happens because \(\frac{k}{k+1}\) approaches 1, and the natural logarithm of 1 is zero. Hence within series, it often serves to reduce the size or order of terms, making them more comparable to known functions like the \(\frac{1}{k+1}\) in the comparison.

Harmonic Series

The harmonic series is a classic series in mathematics given by \(\sum_{k=1}^{\infty} \frac{1}{k}\). It shows up frequently in various convergence and divergence discussions.

Despite how slowly the terms get smaller, the harmonic series diverges. This means that if we try to sum all these \(\frac{1}{k}\) terms from 1 to infinity, the sum grows without bound. However, if we consider a modified version like \(\sum_{k=1}^{\infty} \frac{1}{k+1}\), which slightly shifts the sequence, the nature remains the same, diverging as well.

In our series convergence problem, this property is crucial. By matching terms like \(p \ln\left(\frac{k}{k+1}\right)\) to a known translated harmonic series, we could apply the Comparison Test to great effect, revealing that our series converges for values of \(p > 1\), leveraging the divergent nature of the harmonic series with a specific value of \(p\). This example demonstrates the harmonic series's pivotal role in determining the convergence or divergence behavior of related series.