Problem 16 - Calculus: Early Transcendentals

Question

Test the series for convergence or divergence. \( \displaystyle \sum_{n = 1}^{\infty} \frac {\sqrt{n^4 + 1}}{n^3 + n} \)

Step-by-Step Solution

Answer

The series is divergent by the Limit Comparison Test.

Step 1: Simplify the series term

First, let's analyze the general term of the series: \( a_n = \frac{\sqrt{n^4 + 1}}{n^3 + n} \). For large \( n \), the dominant term in the square root \( \sqrt{n^4 + 1} \) is \( n^2 \), and in the denominator \( n^3 + n \), the dominant term is \( n^3 \). Thus, for large \( n \), \( a_n \approx \frac{n^2}{n^3} = \frac{1}{n} \).

Step 2: Use the Comparison Test

We compare \( a_n = \frac{\sqrt{n^4 + 1}}{n^3 + n} \) to \( b_n = \frac{1}{n} \). We observed that as \( n \to \infty \), \( a_n \approx b_n = \frac{1}{n} \). Since \( \sum \frac{1}{n} \) (the harmonic series) is divergent, we might conjecture that \( \sum a_n \) could also be divergent. However, for the Comparison Test, since the terms \( a_n \leq b_n \) for all sufficiently large \( n \), and given that \( b_n \) is divergent, this test is inconclusive as we didn’t satisfy \( a_n \geq b_n \).

Step 3: Use the Limit Comparison Test

Let's apply the Limit Comparison Test. We compute \( \lim_{{n \to \infty}} \frac{a_n}{b_n} = \lim_{{n \to \infty}} \frac{\sqrt{n^4+1}}{n^3+n} \cdot n = \lim_{{n \to \infty}} \frac{\sqrt{n^4+1}}{n^2+n^0}. \) Simplifying inside the limit: which is \( \lim_{{n \to \infty}} \frac{n^2\sqrt{1+\frac{1}{n^4}}}{1+\frac{1}{n^2}} = \lim_{{n \to \infty}} \frac{n^2}{1} \cdot \frac{\sqrt{1+0}}{1} = 1 \).

Step 4: Conclude with the Limit Comparison Test

Since \( \lim_{{n \to \infty}} \frac{a_n}{b_n} = 1 \) and \( \sum b_n = \sum \frac{1}{n}\) is a divergent series, by the Limit Comparison Test, \( \sum a_n \) is also divergent.

Key Concepts

Comparison Test

The comparison test is a useful method for determining whether a series converges or diverges. It involves comparing the series in question to another series whose convergence behavior is already known. If you have a series \[ \sum a_n \] and a second series \[ \sum b_n \] where \[ a_n \leq b_n \] for every term in the series, then:

• If \[ \sum b_n \] is convergent, \[ \sum a_n \] is also convergent.
• If \[ \sum a_n \] is divergent, then \[ \sum b_n \] is also divergent.

The trick is to pick a comparator series wisely. If the comparator series are divergent like the harmonic series, and you can only show \[ a_n \leq b_n \], it's inconclusive for convergence. Thus, finding the right comparison is key to the test's success.

Limit Comparison Test

The limit comparison test offers a more nuanced way to determine series divergence or convergence, especially when the comparison test is inconclusive. For two series \[ \sum a_n \] and \[ \sum b_n \], calculate \[ \lim_{{n \to \infty}} \frac{a_n}{b_n} = L. \]

• If \( L \) is a positive finite number, both series will converge or diverge together.
• If \( L = 0 \) and \[ \sum b_n \] converges, then \[ \sum a_n \] converges too.
• Conversely, if \( L = \infty \) and \[ \sum b_n \] is divergent, then \[ \sum a_n \] is divergent.

The benefit of this test is that it provides a direct relationship between the two series. It overcomes the limitations encountered in the simple comparison test. In the given exercise, we used this successfully to determine that our series was divergent.

Harmonic Series

The harmonic series is a fundamental concept in understanding series convergence behavior. It is represented by:\[ \sum_{n=1}^{\infty} \frac{1}{n}, \] which is famous because it diverges. Despite each individual term \( \frac{1}{n} \) getting smaller as \( n \) increases, the series doesn't settle into a finite sum.
This aspect of the harmonic series is crucial in comparison and limit comparison tests. It serves as a benchmark for proving divergence in other series. The intuitive reason for divergence is that even though later terms are very small, there are infinitely many of them, ensuring the sum continues growing without bound. Understanding this series is critical for recognizing divergence patterns in more complex series like the one in our exercise.

Divergence of Series

Divergence in series means that the sum does not approach a specific number as more terms are added. Instead, the total grows indefinitely or fluctuates without settling. The divergence signifies that no single value can adequately represent the total sum of all terms in the series.
Testing for divergence often involves methods such as:

• Direct definition analysis (if terms \( a_n \) do not approach zero, the series diverges).
• Comparison tests with known divergent series.
• Ratio or root tests if applicable.

Knowing whether a series converges or diverges is fundamental to many areas of calculus and mathematical analysis. It informs whether attempting to find a sum is worthwhile or which mathematical tools to apply for approximation.