Problem 11 - Calculus: Early Transcendentals

Question

Determine whether the series converges or diverges. \( \displaystyle \sum_{k = 1}^{\infty} \frac { \sqrt [3]{k}}{\sqrt {k^3 + 4k + 3}} \)

Step-by-Step Solution

Answer

The series converges by comparison with a p-series.

Step 1: Analyze the Series Form

The series to determine convergence or divergence is \( \sum_{k = 1}^{\infty} \frac{\sqrt[3]{k}}{\sqrt{k^3 + 4k + 3}} \). It is often helpful to determine the behavior of the function within the series for very large values of \( k \).

Step 2: Simplify the Series Expression at Infinity

For very large \( k \), the term \( k^3 \) in the denominator \( \sqrt{k^3 + 4k + 3} \) dominates. Thus, \( \sqrt{k^3 + 4k + 3} \approx \sqrt{k^3} = k^{3/2} \). Therefore, the individual term \( \frac{\sqrt[3]{k}}{\sqrt{k^3 + 4k + 3}} \approx \frac{k^{1/3}}{k^{3/2}} = \frac{1}{k^{7/6}} \).

Step 3: Compare with p-series

A p-series is of the form \( \sum \frac{1}{k^p} \). The simplified term \( \frac{1}{k^{7/6}} \) suggests comparing the original series to a p-series with \( p = \frac{7}{6} \). A p-series converges if \( p > 1 \), and in this case, \( \frac{7}{6} > 1 \).

Step 4: Conclude Using the Comparison Test

Using the comparison test, we find that since \( \sum \frac{1}{k^{7/6}} \) converges, and \( \frac{\sqrt[3]{k}}{\sqrt{k^3 + 4k + 3}} \approx \frac{1}{k^{7/6}} \) for large \( k \), the original series \( \sum_{k=1}^{\infty} \frac{\sqrt[3]{k}}{\sqrt{k^3 + 4k + 3}} \) also converges.

Key Concepts

Comparison Test

The comparison test is a powerful tool in determining the convergence or divergence of series. It involves comparing a given series to another series whose convergence behavior is already known. Here's how it works:

• If a series \(\sum a_k\) is such that \(a_k\) is always less than or equal to \(b_k\) for all \(k\) and \(\sum b_k\) converges, then \(\sum a_k\) also converges.
• On the other hand, if \(a_k\) is always greater than or equal to \(b_k\) and \(\sum b_k\) diverges, then \(\sum a_k\) diverges too.

By using the comparison test, we can often establish the convergence or divergence of complex series by relating them to simpler, more familiar series.

P-Series

A p-series is a specific type of series that takes the form \(\sum \frac{1}{k^p}\). Understanding p-series helps in comparison because their convergence is well-defined:

• A p-series converges if and only if \(p > 1\).
• If \(p = 1\), the series becomes the harmonic series \(\sum \frac{1}{k}\), which is known to diverge.

In the context of the original exercise, once the given series is simplified, it results in something very similar to a p-series. Specifically, it compares to a series where \(p = \frac{7}{6}\), which is greater than 1, providing a straightforward path to deducing convergence.

Dominant Term Analysis

Dominant term analysis simplifies series by focusing on the terms that influence its behavior the most, especially as \(k\) approaches infinity. For the given series \(\sum_{k=1}^{\infty} \frac{\sqrt[3]{k}}{\sqrt{k^3 + 4k + 3}}\), the denominator \(\sqrt{k^3 + 4k + 3}\) can be approximated to \(\sqrt{k^3}\) because as \(k\) becomes large, \(k^3\) overshadows the other terms.

• This simplification leads us to the term \(\frac{\sqrt[3]{k}}{\sqrt{k^3}} = \frac{k^{1/3}}{k^{3/2}} = \frac{1}{k^{7/6}}\).
• The dominant terms in the polynomial indicates the series behaves similarly to a p-series, making it easier to analyze.

Employing dominant term analysis allows for reducing complex series into simpler forms equivalent to known series, facilitating easier convergence determination.