Problem 43 - Calculus: Early Transcendentals

Question

For which of the following series is the Ratio Test inconclusive (that is, if fails to give a definite answer)? (a) \( \displaystyle \sum_{n = 1}^{\infty} \frac {1}{n^3} \) (b) \( \displaystyle \sum_{n = 1}^{\infty} \frac {n}{n^2} \) (c) \( \displaystyle \sum_{n = 1}^{\infty} \frac {( - 3)^{n-1}}{\sqrt{n}} \) (d) \( \displaystyle \sum_{n = 1}^{\infty} \frac {\sqrt{n}}{1 + n^2} \)

Step-by-Step Solution

Answer

The Ratio Test is inconclusive for series (a), (b), and (d).

Step 1: Apply Ratio Test to Series (a)

The series given is \( \sum_{n=1}^{\infty} \frac{1}{n^3} \). Calculate the ratio \( \frac{a_{n+1}}{a_n} \), where \( a_n = \frac{1}{n^3} \). Thus, \( a_{n+1} = \frac{1}{(n+1)^3} \). The ratio is \( \frac{a_{n+1}}{a_n} = \frac{n^3}{(n+1)^3} \). As \( n \to \infty \), the limit becomes \( \left(\frac{n}{n+1}\right)^3 \to 1^3 = 1 \). This indicates the Ratio Test is inconclusive.

Step 2: Apply Ratio Test to Series (b)

The series given is \( \sum_{n=1}^{\infty} \frac{n}{n^2} \). Simplifying, this becomes \( \sum_{n=1}^{\infty} \frac{1}{n} \). The terms are \( a_n = \frac{1}{n} \) and \( a_{n+1} = \frac{1}{n+1} \). The ratio \( \frac{a_{n+1}}{a_n} = \frac{n}{n+1} \). As \( n \to \infty \), the limit is \( 1 \), indicating the Ratio Test is inconclusive.

Step 3: Apply Ratio Test to Series (c)

The series given is \( \sum_{n=1}^{\infty} \frac{(-3)^{n-1}}{\sqrt{n}} \). Set \( a_n = \frac{(-3)^{n-1}}{\sqrt{n}} \) and \( a_{n+1} = \frac{(-3)^n}{\sqrt{n+1}} \). Compute \( \frac{a_{n+1}}{a_n} = \frac{(-3)^n\sqrt{n}}{(-3)^{n-1}\sqrt{n+1}} = -3 \cdot \frac{\sqrt{n}}{\sqrt{n+1}} \). The limit is \(-3 \cdot 1 = -3 \), as \( n \to \infty \). This value shows the series diverges by absolute value, so the Ratio Test is conclusive.

Step 4: Apply Ratio Test to Series (d)

The series given is \( \sum_{n=1}^{\infty} \frac{\sqrt{n}}{1+n^2} \). Set \( a_n = \frac{\sqrt{n}}{1+n^2} \) and \( a_{n+1} = \frac{\sqrt{n+1}}{1+(n+1)^2} \). Compute \( \frac{a_{n+1}}{a_n} = \frac{\sqrt{n+1}}{\sqrt{n}} \cdot \frac{1+n^2}{1+(n+1)^2} \). Simplifying yields the limit \( \frac{\sqrt{n+1}}{\sqrt{n}} \to 1 \) and \( \frac{1+n^2}{1+(n+1)^2} \to 1 \), so the product is \( 1 \). The Ratio Test is inconclusive for this series.

Key Concepts

Convergence of Series

The concept of convergence is foundational when discussing infinite series. An infinite series converges if the sum of its terms approaches a finite limit as more terms are added. Specifically, for a series \( \sum a_n \), if the sequence of partial sums \( S_n = a_1 + a_2 + \ldots + a_n \) approaches a specific number \( L \) as \( n \to \infty \), the series converges to \( L \).
Convergence is crucial in determining if the infinite addition of terms makes sense mathematically or leads to a specific sum.

• A famous example of a convergent series is the geometric series \( \sum_{n=0}^{\infty} ar^n \) for \(|r| < 1\).
• Convergence ensures that adding infinitely many terms does not lead to infinity but rather to a specific value.

Understanding convergence helps in a wide array of mathematical applications, from solving differential equations to analyzing financial models.

Divergence of Series

Divergence is the counterpart to convergence. An infinite series diverges if the sum of its terms does not approach a finite limit. This means that either the series' terms do not trend towards zero, or their partial sums do not stabilize at a single number.
For example, the harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \) diverges because its terms tend too slowly towards zero. Although the individual terms diminish, their cumulative sum increases without bound.

• Determining divergence is essential to avoid misrepresentations in calculations and ensure accurate mathematical modeling.
• Divergence indicates that the series does not fit within our usual understanding of finite addition.

Methods such as the Divergence Test, which checks if the limit of the terms themselves is non-zero, often help identify divergent series.

Limit Comparison Test

The Limit Comparison Test is a helpful tool in the analysis of series to check for convergence or divergence. It compares the behavior of two series to draw conclusions about convergence.
To apply the test, you select a known benchmark series, say \( \sum b_n \), and compare it with the series in question \( \sum a_n \).
The test states:

• If \( \lim_{n \to \infty} \frac{a_n}{b_n} = c \), where \( 0 < c < \infty \), then both series converge or both diverge.
• This test is especially useful when the series is complicated, but can be compared to a simpler series whose behavior is known.

The Limit Comparison Test facilitates understanding by relating complex series to simpler, well-understood series, offering clarity on their convergence or divergence.