Problem 12 - Essential Calculus Early Transcendentals

Question

Determine whether the series is convergent or divergent. $$\sum_{n=1}^{\infty}\left(n^{-1.4}+3 n^{-1.2}\right)$$

Step-by-Step Solution

Answer

The series is convergent.

Step 1: Identify the series components

The given series is \( \sum_{n=1}^{\infty} \left( n^{-1.4} + 3n^{-1.2} \right) \). This series is the sum of two separate series: \( \sum_{n=1}^{\infty} n^{-1.4} \) and \( \sum_{n=1}^{\infty} 3n^{-1.2} \).

Step 2: Apply the p-series test

Both series can be considered p-series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p = 1.4 \) and \( p = 1.2 \) respectively. According to the p-series test, a p-series converges if \( p > 1 \).

Step 3: Determine convergence of each component

For the first series, \( \sum_{n=1}^{\infty} n^{-1.4} \), \( p = 1.4 > 1 \), so this series converges. For the second series, \( \sum_{n=1}^{\infty} 3n^{-1.2} \), \( p = 1.2 > 1 \), so this series also converges.

Step 4: Conclude about the original series

Since both series \( \sum_{n=1}^{\infty} n^{-1.4} \) and \( \sum_{n=1}^{\infty} 3n^{-1.2} \) converge individually, the entire series \( \sum_{n=1}^{\infty} \left( n^{-1.4} + 3n^{-1.2} \right) \) also converges by the sum of convergent series property.

Key Concepts

p-series test

The **p-series test** is a fundamental tool for determining the convergence of infinite series. An infinite series takes the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive constant. The convergence of these series is determined by the value of \( p \).For a p-series to converge:

• \( p > 1 \): This implies that the terms of the series are getting smaller quickly enough for the series to converge.
• \( p \leq 1 \): In contrast, if \( p \) is less than or equal to 1, the series diverges. This indicates that the terms are not decreasing rapidly enough to sum to a finite value.

In the example, we have two p-series components: \( n^{-1.4} \) and \( n^{-1.2} \). Both have values of \( p \) greater than 1, confirming the convergence of each series individually.

sum of convergent series

Understanding the **sum of convergent series** is crucial when dealing with complex series that can be broken down into simpler parts. If you have multiple series that each converge, the sum of these series will also converge.Let's consider how this property applies:

• If \( \sum a_n \) and \( \sum b_n \) are both convergent, then \( \sum (a_n + b_n) \) is convergent too.
• The convergence of the entire series follows from the individual convergence of each part.

In the given exercise, the series \( \sum_{n=1}^{\infty} \left( n^{-1.4} + 3n^{-1.2} \right) \) can be split as:

• \( \sum_{n=1}^{\infty} n^{-1.4} \)
• \( \sum_{n=1}^{\infty} 3n^{-1.2} \)

Both these series are convergent, according to the p-series test, ensuring that the sum of them is also convergent.

convergence criteria

**Convergence criteria** help determine whether a series will result in a finite sum. These criteria are essential for identifying the behavior of a series. Several tests can be applied to decide whether a series converges or diverges, such as the p-series test, comparison test, ratio test, and others.In this context, when dealing with series:

• Assessing the nature and rate of decay of the series terms is crucial.
• Using the right convergence test depends on the form and complexity of the series.

For p-series, the criteria are specifically about comparing the exponent \( p \) to 1. When \( p \) is greater than 1, each term in the series shrinks swiftly enough, leading to a convergent series. The exercise focuses on this principle by utilizing the p-series test to evaluate convergence.By applying the appropriate criteria, such as the p-series test in this instance, we gain insight into the series' behavior, ensuring a comprehensive understanding of why and how a series converges.