Question

In the following exercises, use an appropriate test to determine whether the series converges. $$ \left.a_{k}=\left(\frac{k}{k+\ln k}\right)^{k} \text { (Hint: } a_{k}=\left(1+\frac{\ln k}{k}\right)^{-(k / \ln k) \ln k} \approx e^{-\ln k} \cdot\right) $$

Short answer

The series diverges.

Step-by-step solution

Understand the Series

We are given the series \( a_k = \left(\frac{k}{k+\ln k}\right)^k \). The hint suggests rewriting \( a_k \) and analyzing its approximate behavior.

Rewrite the Series Using Approximation

Rewrite the term using the hint: \( a_k = \left(1 + \frac{\ln k}{k}\right)^{-\left(\frac{k}{\ln k}\right)\ln k} \approx e^{-\ln k} \). This equals \( \frac{1}{k} \).

Compare with Harmonic Series

Notice that \( \frac{1}{k} \) represents the harmonic series, which is known to diverge. This serves as an important comparison for our original series terms, suggesting potential divergence.

Apply Limit Comparison or Other Convergence Tests

Use the Limit Comparison Test with the harmonic series \( \sum \frac{1}{k} \). Calculate the limit: \( \lim_{k \to \infty} \frac{a_k}{\frac{1}{k}} = \lim_{k \to \infty} k a_k \). If the limit is finite and non-zero, both series either converge or diverge together.

Calculate the Limit

Substitute \( a_k = \left(\frac{k}{k+\ln k}\right)^k \) into the limit: \[ \lim_{k \to \infty} k \left(\frac{k}{k+\ln k}\right)^k = \lim_{k \to \infty} \left(\frac{k}{k+\ln k}\right)^k = e^{-1} \] since the inner approximation is \( e^{-\ln k} \).

Conclusion from Limit Result

The limit calculation revealed that the \( \lim_{k \to \infty} k a_k = e^{-1} \), a finite and non-zero number. Therefore, by the Limit Comparison Test, since the harmonic series \( \sum \frac{1}{k} \) diverges, the given series must also diverge.

Key concepts

Limit Comparison Test

The Limit Comparison Test is a powerful technique for determining the convergence or divergence of series. It works by comparing the series in question with another series whose behavior is well-known. This is done by calculating the limit of the ratio of their terms. Here's how it works:

• Select another series, which we suspect behaves similarly to the original series. Usually, this is a series whose convergence or divergence is already established.
• Calculate the limit of the ratio of their terms as the index approaches infinity: \[ \lim_{k \to \infty} \frac{a_k}{b_k} \]
• If the limit is a positive finite number, both series either converge or diverge together.

For instance, in the exercise given, the series term \( a_k = \left(\frac{k}{k+\ln k}\right)^k \) is compared to the harmonic series \( \frac{1}{k} \). After computing the limit and finding it to be \( e^{-1} \), which is positive and finite, we conclude that since the harmonic series diverges, so does the given series.

Harmonic Series

The harmonic series is one of the simplest examples of an infinite series, expressed as \( \sum_{k=1}^{\infty} \frac{1}{k} \). Each term in the harmonic series is the reciprocal of an integer. Although the terms get smaller as \( k \) increases, surprisingly, the series diverges.Here's a quick breakdown of why the harmonic series diverges:

• The terms decrease in size but not rapidly enough for the sum to converge to a finite value.
• One way to understand this is through the integral test. The integral of \( \frac{1}{x} \) from 1 to infinity is infinite, showing that the harmonic series also diverges.

In the exercise, we use the harmonic series as a benchmark in the Limit Comparison Test to determine whether the original series diverges.

Divergence of Series

Divergence of a series means that the sum of the series is infinite or does not approach a finite number as more terms are added. Understanding divergence is crucial when analyzing series because it tells us when a series does not have a sum in the traditional sense. Key points about divergence:

• If a series diverges, no matter how large the number of terms summed, the total will grow without bounds or fluctuate indefinitely.
• Recognizing familiar diverging series, like the harmonic series, helps in determining the behavior of similar series using comparison tests such as the Limit Comparison Test.

In our exercise, the series was shown to diverge by using the Limit Comparison Test with the diverging harmonic series, hence confirming that the given series lacks convergence.